#### 问题：浮点数学运算是否被破坏？

``````0.1 + 0.2 == 0.3  ->  false
``````
``````0.1 + 0.2         ->  0.30000000000000004
``````

#### Q: Is floating point math broken?

Consider the following code:

``````0.1 + 0.2 == 0.3  ->  false
``````
``````0.1 + 0.2         ->  0.30000000000000004
``````

Why do these inaccuracies happen?

#### 回答1：

• `0.1`(十进制)或
• `0x1.99999999999999...p-4`(类似于C99十六进制表示法)，其中`...`表示无休止的9序列。

**当然，这并不完全是将浮点数存储在内存中的方式(它们使用科学计数形式)。但是，它确实说明了二进制浮点精度误差趋于增加的观点，因为我们通常感兴趣的"真实世界"数字通常是10的幂-但这仅仅是因为我们使用了十进制数天-今天。这也是为什么我们要说71％而不是"每7个中有5个"(71％是一个近似值，因为5/7不能用任何十进制数字精确表示)的原因。

，请

#### A1:

Binary floating point math is like this. In most programming languages, it is based on the IEEE 754 standard. The crux of the problem is that numbers are represented in this format as a whole number times a power of two; rational numbers (such as `0.1`, which is `1/10`) whose denominator is not a power of two cannot be exactly represented.

For `0.1` in the standard `binary64` format, the representation can be written exactly as

• `0.1000000000000000055511151231257827021181583404541015625` in decimal, or
• `0x1.999999999999ap-4` in C99 hexfloat notation.

In contrast, the rational number `0.1`, which is `1/10`, can be written exactly as

• `0.1` in decimal, or
• `0x1.99999999999999...p-4` in an analogue of C99 hexfloat notation, where the `...` represents an unending sequence of 9's.

The constants `0.2` and `0.3` in your program will also be approximations to their true values. It happens that the closest `double` to `0.2` is larger than the rational number `0.2` but that the closest `double` to `0.3` is smaller than the rational number `0.3`. The sum of `0.1` and `0.2` winds up being larger than the rational number `0.3` and hence disagreeing with the constant in your code.

A fairly comprehensive treatment of floating-point arithmetic issues is What Every Computer Scientist Should Know About Floating-Point Arithmetic. For an easier-to-digest explanation, see floating-point-gui.de.

Side Note: All positional (base-N) number systems share this problem with precision

Plain old decimal (base 10) numbers have the same issues, which is why numbers like 1/3 end up as 0.333333333...

You've just stumbled on a number (3/10) that happens to be easy to represent with the decimal system, but doesn't fit the binary system. It goes both ways (to some small degree) as well: 1/16 is an ugly number in decimal (0.0625), but in binary it looks as neat as a 10,000th does in decimal (0.0001)** - if we were in the habit of using a base-2 number system in our daily lives, you'd even look at that number and instinctively understand you could arrive there by halving something, halving it again, and again and again.

** Of course, that's not exactly how floating-point numbers are stored in memory (they use a form of scientific notation). However, it does illustrate the point that binary floating-point precision errors tend to crop up because the "real world" numbers we are usually interested in working with are so often powers of ten - but only because we use a decimal number system day-to-day. This is also why we'll say things like 71% instead of "5 out of every 7" (71% is an approximation, since 5/7 can't be represented exactly with any decimal number).

So no: binary floating point numbers are not broken, they just happen to be as imperfect as every other base-N number system :)

Side Side Note: Working with Floats in Programming

In practice, this problem of precision means you need to use rounding functions to round your floating point numbers off to however many decimal places you're interested in before you display them.

You also need to replace equality tests with comparisons that allow some amount of tolerance, which means:

Do not do `if (x == y) { ... }`

Instead do `if (abs(x - y) < myToleranceValue) { ... }`.

where `abs` is the absolute value. `myToleranceValue` needs to be chosen for your particular application - and it will have a lot to do with how much "wiggle room" you are prepared to allow, and what the largest number you are going to be comparing may be (due to loss of precision issues). Beware of "epsilon" style constants in your language of choice. These are not to be used as tolerance values.

3.1除法舍入误差：倒数的近似值

# A Hardware Designer's Perspective

I believe I should add a hardware designer’s perspective to this since I design and build floating point hardware. Knowing the origin of the error may help in understanding what is happening in the software, and ultimately, I hope this helps explain the reasons for why floating point errors happen and seem to accumulate over time.

## 1. Overview

From an engineering perspective, most floating point operations will have some element of error since the hardware that does the floating point computations is only required to have an error of less than one half of one unit in the last place. Therefore, much hardware will stop at a precision that's only necessary to yield an error of less than one half of one unit in the last place for a single operation which is especially problematic in floating point division. What constitutes a single operation depends upon how many operands the unit takes. For most, it is two, but some units take 3 or more operands. Because of this, there is no guarantee that repeated operations will result in a desirable error since the errors add up over time.

## 2. Standards

Most processors follow the IEEE-754 standard but some use denormalized, or different standards . For example, there is a denormalized mode in IEEE-754 which allows representation of very small floating point numbers at the expense of precision. The following, however, will cover the normalized mode of IEEE-754 which is the typical mode of operation.

In the IEEE-754 standard, hardware designers are allowed any value of error/epsilon as long as it's less than one half of one unit in the last place, and the result only has to be less than one half of one unit in the last place for one operation. This explains why when there are repeated operations, the errors add up. For IEEE-754 double precision, this is the 54th bit, since 53 bits are used to represent the numeric part (normalized), also called the mantissa, of the floating point number (e.g. the 5.3 in 5.3e5). The next sections go into more detail on the causes of hardware error on various floating point operations.

## 3. Cause of Rounding Error in Division

The main cause of the error in floating point division is the division algorithms used to calculate the quotient. Most computer systems calculate division using multiplication by an inverse, mainly in `Z=X/Y`, `Z = X * (1/Y)`. A division is computed iteratively i.e. each cycle computes some bits of the quotient until the desired precision is reached, which for IEEE-754 is anything with an error of less than one unit in the last place. The table of reciprocals of Y (1/Y) is known as the quotient selection table (QST) in the slow division, and the size in bits of the quotient selection table is usually the width of the radix, or a number of bits of the quotient computed in each iteration, plus a few guard bits. For the IEEE-754 standard, double precision (64-bit), it would be the size of the radix of the divider, plus a few guard bits k, where `k>=2`. So for example, a typical Quotient Selection Table for a divider that computes 2 bits of the quotient at a time (radix 4) would be `2+2= 4` bits (plus a few optional bits).

3.1 Division Rounding Error: Approximation of Reciprocal

What reciprocals are in the quotient selection table depend on the division method: slow division such as SRT division, or fast division such as Goldschmidt division; each entry is modified according to the division algorithm in an attempt to yield the lowest possible error. In any case, though, all reciprocals are approximations of the actual reciprocal and introduce some element of error. Both slow division and fast division methods calculate the quotient iteratively, i.e. some number of bits of the quotient are calculated each step, then the result is subtracted from the dividend, and the divider repeats the steps until the error is less than one half of one unit in the last place. Slow division methods calculate a fixed number of digits of the quotient in each step and are usually less expensive to build, and fast division methods calculate a variable number of digits per step and are usually more expensive to build. The most important part of the division methods is that most of them rely upon repeated multiplication by an approximation of a reciprocal, so they are prone to error.

## 4. Rounding Errors in Other Operations: Truncation

Another cause of the rounding errors in all operations are the different modes of truncation of the final answer that IEEE-754 allows. There's truncate, round-towards-zero, round-to-nearest (default), round-down, and round-up. All methods introduce an element of error of less than one unit in the last place for a single operation. Over time and repeated operations, truncation also adds cumulatively to the resultant error. This truncation error is especially problematic in exponentiation, which involves some form of repeated multiplication.

## 5. Repeated Operations

Since the hardware that does the floating point calculations only needs to yield a result with an error of less than one half of one unit in the last place for a single operation, the error will grow over repeated operations if not watched. This is the reason that in computations that require a bounded error, mathematicians use methods such as using the round-to-nearest even digit in the last place of IEEE-754, because, over time, the errors are more likely to cancel each other out, and Interval Arithmetic combined with variations of the IEEE 754 rounding modes to predict rounding errors, and correct them. Because of its low relative error compared to other rounding modes, round to nearest even digit (in the last place), is the default rounding mode of IEEE-754.

Note that the default rounding mode, round-to-nearest even digit in the last place, guarantees an error of less than one half of one unit in the last place for one operation. Using the truncation, round-up, and round down alone may result in an error that is greater than one half of one unit in the last place, but less than one unit in the last place, so these modes are not recommended unless they are used in Interval Arithmetic.

## 6. Summary

In short, the fundamental reason for the errors in floating point operations is a combination of the truncation in hardware, and the truncation of a reciprocal in the case of division. Since the IEEE-754 standard only requires an error of less than one half of one unit in the last place for a single operation, the floating point errors over repeated operations will add up unless corrected.

#### A3:

When you convert .1 or 1/10 to base 2 (binary) you get a repeating pattern after the decimal point, just like trying to represent 1/3 in base 10. The value is not exact, and therefore you can't do exact math with it using normal floating point methods.

#### 回答4：

(这两个数字之间的差是我们必须决定包括在内的"最小切片"，这会导致向上的偏差，或者排除在外的情况会导致向下的偏差。最小切片的技术术语是 ulp 。)

P.S。某些编程语言还提供了比萨饼切割器，可以将切片切成十分之一。尽管这种披萨切刀并不常见，但是如果您确实有机会使用它，那么在重要的是要精确获得十分之一或五分之一的切片时，应该使用它。

#### A4:

Most answers here address this question in very dry, technical terms. I'd like to address this in terms that normal human beings can understand.

Imagine that you are trying to slice up pizzas. You have a robotic pizza cutter that can cut pizza slices exactly in half. It can halve a whole pizza, or it can halve an existing slice, but in any case, the halving is always exact.

That pizza cutter has very fine movements, and if you start with a whole pizza, then halve that, and continue halving the smallest slice each time, you can do the halving 53 times before the slice is too small for even its high-precision abilities. At that point, you can no longer halve that very thin slice, but must either include or exclude it as is.

Now, how would you piece all the slices in such a way that would add up to one-tenth (0.1) or one-fifth (0.2) of a pizza? Really think about it, and try working it out. You can even try to use a real pizza, if you have a mythical precision pizza cutter at hand. :-)

Most experienced programmers, of course, know the real answer, which is that there is no way to piece together an exact tenth or fifth of the pizza using those slices, no matter how finely you slice them. You can do a pretty good approximation, and if you add up the approximation of 0.1 with the approximation of 0.2, you get a pretty good approximation of 0.3, but it's still just that, an approximation.

For double-precision numbers (which is the precision that allows you to halve your pizza 53 times), the numbers immediately less and greater than 0.1 are 0.09999999999999999167332731531132594682276248931884765625 and 0.1000000000000000055511151231257827021181583404541015625. The latter is quite a bit closer to 0.1 than the former, so a numeric parser will, given an input of 0.1, favour the latter.

(The difference between those two numbers is the "smallest slice" that we must decide to either include, which introduces an upward bias, or exclude, which introduces a downward bias. The technical term for that smallest slice is an ulp.)

In the case of 0.2, the numbers are all the same, just scaled up by a factor of 2. Again, we favour the value that's slightly higher than 0.2.

Notice that in both cases, the approximations for 0.1 and 0.2 have a slight upward bias. If we add enough of these biases in, they will push the number further and further away from what we want, and in fact, in the case of 0.1 + 0.2, the bias is high enough that the resulting number is no longer the closest number to 0.3.

In particular, 0.1 + 0.2 is really 0.1000000000000000055511151231257827021181583404541015625 + 0.200000000000000011102230246251565404236316680908203125 = 0.3000000000000000444089209850062616169452667236328125, whereas the number closest to 0.3 is actually 0.299999999999999988897769753748434595763683319091796875.

P.S. Some programming languages also provide pizza cutters that can split slices into exact tenths. Although such pizza cutters are uncommon, if you do have access to one, you should use it when it's important to be able to get exactly one-tenth or one-fifth of a slice.

(Originally posted on Quora.)

#### A5:

Floating point rounding errors. 0.1 cannot be represented as accurately in base-2 as in base-10 due to the missing prime factor of 5. Just as 1/3 takes an infinite number of digits to represent in decimal, but is "0.1" in base-3, 0.1 takes an infinite number of digits in base-2 where it does not in base-10. And computers don't have an infinite amount of memory.

#### 回答6：

``````var result = 1.0 + 2.0;     // result === 3.0 returns true
``````

...而不是：

``````var result = 0.1 + 0.2;     // result === 0.3 returns false
``````

1 道格拉斯·克罗克福德："> JavaScript：好的部分：附录A-糟糕的部分(第105页)

#### A6:

In addition to the other correct answers, you may want to consider scaling your values to avoid problems with floating-point arithmetic.

For example:

``````var result = 1.0 + 2.0;     // result === 3.0 returns true
``````

``````var result = 0.1 + 0.2;     // result === 0.3 returns false
``````

The expression `0.1 + 0.2 === 0.3` returns `false` in JavaScript, but fortunately integer arithmetic in floating-point is exact, so decimal representation errors can be avoided by scaling.

As a practical example, to avoid floating-point problems where accuracy is paramount, it is recommended1 to handle money as an integer representing the number of cents: `2550` cents instead of `25.50` dollars.

1 Douglas Crockford: JavaScript: The Good Parts: Appendix A - Awful Parts (page 105).

#### 回答7：

IEEE 754双精度二进制浮点格式(binary64)数字表示形式的数字

value =(-1)^ s *(1.m 51 m 50 ... m 2 m 1 m 0 ) 2 * 2 e-1023

64位：

• 第一位是签名位`1`(如果数字为负，`0`否则为 1
• 接下来的11位是指数，即偏移量乘以1023。换句话说，从双精度数中读取指数位后，必须减去1023才能获得两个。
• 剩余的52位是有效数(或尾数)。在尾数中，"隐含的" `1.`总是 2 被省略，因为任何二进制值的最高有效位是`1`

1 -IEEE 754允许使用签名零的概念-`+0``-0`的区别对待：`1//(+0)`是正无穷大； `1/(-0)`是负无穷大。对于零值，尾数和指数位均为零。注意：零值(+0和-0)明确未归类为denormal 2

2 -反常数并非如此，偏移指数为零(隐含的`0.`)。异常双精度数的范围是d min ≤| x | ≤d max ，其中d min (最小的可表示非零数字)为2 -1023-51 (≈4.94 * 10 - 324 )和d max (最大的非正规数，尾数完全由`1` s组成)是2 -1023 + 1 -2 -1023-51 (≈2.225 * 10 -308 )。

``````public static string BinaryRepresentation(double value)
{
long valueInLongType = BitConverter.DoubleToInt64Bits(value);
string bits = Convert.ToString(valueInLongType, 2);
string leadingZeros = new string('0', 64 - bits.Length);
string binaryRepresentation = leadingZeros + bits;

string sign = binaryRepresentation.ToString();
string exponent = binaryRepresentation.Substring(1, 11);
string mantissa = binaryRepresentation.Substring(12);

return string.Format("{0}:{1}:{2}", sign, exponent, mantissa);
}
``````

(跳至TL； DR版本的底部)

IEEE 754以二进制形式(用冒号分隔三个部分)编写，其值表示为：

``````0.1 => 0:01111111011:1001100110011001100110011001100110011001100110011010
0.2 => 0:01111111100:1001100110011001100110011001100110011001100110011010
``````

``````0.1 => 2^-4 * .1001100110011001100110011001100110011001100110011010
0.2 => 2^-3 * .1001100110011001100110011001100110011001100110011010
or
0.1 => 2^-56 * 7205759403792794 = 0.1000000000000000055511151231257827021181583404541015625
0.2 => 2^-55 * 7205759403792794 = 0.200000000000000011102230246251565404236316680908203125
``````

``````0.1 => 2^-3 *  0.1100110011001100110011001100110011001100110011001101(0)
0.2 => 2^-3 *  1.1001100110011001100110011001100110011001100110011010
sum =  2^-3 * 10.0110011001100110011001100110011001100110011001100111
or
0.1 => 2^-55 * 3602879701896397  = 0.1000000000000000055511151231257827021181583404541015625
0.2 => 2^-55 * 7205759403792794  = 0.200000000000000011102230246251565404236316680908203125
sum =  2^-55 * 10808639105689191 = 0.3000000000000000166533453693773481063544750213623046875
``````

``````sum = 2^-2  * 1.0011001100110011001100110011001100110011001100110011(1)
= 2^-54 * 5404319552844595.5 = 0.3000000000000000166533453693773481063544750213623046875
``````

``````a = 2^-54 * 5404319552844595 = 0.299999999999999988897769753748434595763683319091796875
= 2^-2  * 1.0011001100110011001100110011001100110011001100110011

x = 2^-2  * 1.0011001100110011001100110011001100110011001100110011(1)

b = 2^-2  * 1.0011001100110011001100110011001100110011001100110100
= 2^-54 * 5404319552844596 = 0.3000000000000000444089209850062616169452667236328125
``````

``````sum = 2^-2  * 1.0011001100110011001100110011001100110011001100110100
= 2^-54 * 5404319552844596 = 0.3000000000000000444089209850062616169452667236328125
``````

``````0.3 => 2^-2  * 1.0011001100110011001100110011001100110011001100110011
=  2^-54 * 5404319552844595 = 0.299999999999999988897769753748434595763683319091796875
``````

0.1和0.2的二进制表示形式是IEEE 754允许的数字的最准确的表示形式。由于默认的舍入模式，这些表示形式的加法得出的值仅不同最低有效位。

TL; DR

``````0.1 + 0.2 => 0:01111111101:0011001100110011001100110011001100110011001100110
0.3       => 0:01111111101:0011001100110011001100110011001100110011001100110
``````

``````0.1 + 0.2 => 0.300000000000000044408920985006...
0.3       => 0.299999999999999988897769753748...
``````

#### A7:

My answer is quite long, so I've split it into three sections. Since the question is about floating point mathematics, I've put the emphasis on what the machine actually does. I've also made it specific to double (64 bit) precision, but the argument applies equally to any floating point arithmetic.

Preamble

An IEEE 754 double-precision binary floating-point format (binary64) number represents a number of the form

value = (-1)^s * (1.m51m50...m2m1m0)2 * 2e-1023

in 64 bits:

• The first bit is the sign bit: `1` if the number is negative, `0` otherwise1.
• The next 11 bits are the exponent, which is offset by 1023. In other words, after reading the exponent bits from a double-precision number, 1023 must be subtracted to obtain the power of two.
• The remaining 52 bits are the significand (or mantissa). In the mantissa, an 'implied' `1.` is always2 omitted since the most significant bit of any binary value is `1`.

1 - IEEE 754 allows for the concept of a signed zero - `+0` and `-0` are treated differently: `1 / (+0)` is positive infinity; `1 / (-0)` is negative infinity. For zero values, the mantissa and exponent bits are all zero. Note: zero values (+0 and -0) are explicitly not classed as denormal2.

2 - This is not the case for denormal numbers, which have an offset exponent of zero (and an implied `0.`). The range of denormal double precision numbers is dmin ≤ |x| ≤ dmax, where dmin (the smallest representable nonzero number) is 2-1023 - 51 (≈ 4.94 * 10-324) and dmax (the largest denormal number, for which the mantissa consists entirely of `1`s) is 2-1023 + 1 - 2-1023 - 51 (≈ 2.225 * 10-308).

Turning a double precision number to binary

Many online converters exist to convert a double precision floating point number to binary (e.g. at binaryconvert.com), but here is some sample C# code to obtain the IEEE 754 representation for a double precision number (I separate the three parts with colons (`:`):

``````public static string BinaryRepresentation(double value)
{
long valueInLongType = BitConverter.DoubleToInt64Bits(value);
string bits = Convert.ToString(valueInLongType, 2);
string leadingZeros = new string('0', 64 - bits.Length);
string binaryRepresentation = leadingZeros + bits;

string sign = binaryRepresentation.ToString();
string exponent = binaryRepresentation.Substring(1, 11);
string mantissa = binaryRepresentation.Substring(12);

return string.Format("{0}:{1}:{2}", sign, exponent, mantissa);
}
``````

Getting to the point: the original question

Cato Johnston (the question asker) asked why 0.1 + 0.2 != 0.3.

Written in binary (with colons separating the three parts), the IEEE 754 representations of the values are:

``````0.1 => 0:01111111011:1001100110011001100110011001100110011001100110011010
0.2 => 0:01111111100:1001100110011001100110011001100110011001100110011010
``````

Note that the mantissa is composed of recurring digits of `0011`. This is key to why there is any error to the calculations - 0.1, 0.2 and 0.3 cannot be represented in binary precisely in a finite number of binary bits any more than 1/9, 1/3 or 1/7 can be represented precisely in decimal digits.

Also note that we can decrease the power in the exponent by 52 and shift the point in the binary representation to the right by 52 places (much like 10-3 * 1.23 == 10-5 * 123). This then enables us to represent the binary representation as the exact value that it represents in the form a * 2p. where 'a' is an integer.

Converting the exponents to decimal, removing the offset, and re-adding the implied `1` (in square brackets), 0.1 and 0.2 are:

``````0.1 => 2^-4 * .1001100110011001100110011001100110011001100110011010
0.2 => 2^-3 * .1001100110011001100110011001100110011001100110011010
or
0.1 => 2^-56 * 7205759403792794 = 0.1000000000000000055511151231257827021181583404541015625
0.2 => 2^-55 * 7205759403792794 = 0.200000000000000011102230246251565404236316680908203125
``````

To add two numbers, the exponent needs to be the same, i.e.:

``````0.1 => 2^-3 *  0.1100110011001100110011001100110011001100110011001101(0)
0.2 => 2^-3 *  1.1001100110011001100110011001100110011001100110011010
sum =  2^-3 * 10.0110011001100110011001100110011001100110011001100111
or
0.1 => 2^-55 * 3602879701896397  = 0.1000000000000000055511151231257827021181583404541015625
0.2 => 2^-55 * 7205759403792794  = 0.200000000000000011102230246251565404236316680908203125
sum =  2^-55 * 10808639105689191 = 0.3000000000000000166533453693773481063544750213623046875
``````

Since the sum is not of the form 2n * 1.{bbb} we increase the exponent by one and shift the decimal (binary) point to get:

``````sum = 2^-2  * 1.0011001100110011001100110011001100110011001100110011(1)
= 2^-54 * 5404319552844595.5 = 0.3000000000000000166533453693773481063544750213623046875
``````

There are now 53 bits in the mantissa (the 53rd is in square brackets in the line above). The default rounding mode for IEEE 754 is 'Round to Nearest' - i.e. if a number x falls between two values a and b, the value where the least significant bit is zero is chosen.

``````a = 2^-54 * 5404319552844595 = 0.299999999999999988897769753748434595763683319091796875
= 2^-2  * 1.0011001100110011001100110011001100110011001100110011

x = 2^-2  * 1.0011001100110011001100110011001100110011001100110011(1)

b = 2^-2  * 1.0011001100110011001100110011001100110011001100110100
= 2^-54 * 5404319552844596 = 0.3000000000000000444089209850062616169452667236328125
``````

Note that a and b differ only in the last bit; `...0011` + `1` = `...0100`. In this case, the value with the least significant bit of zero is b, so the sum is:

``````sum = 2^-2  * 1.0011001100110011001100110011001100110011001100110100
= 2^-54 * 5404319552844596 = 0.3000000000000000444089209850062616169452667236328125
``````

whereas the binary representation of 0.3 is:

``````0.3 => 2^-2  * 1.0011001100110011001100110011001100110011001100110011
=  2^-54 * 5404319552844595 = 0.299999999999999988897769753748434595763683319091796875
``````

which only differs from the binary representation of the sum of 0.1 and 0.2 by 2-54.

The binary representation of 0.1 and 0.2 are the most accurate representations of the numbers allowable by IEEE 754. The addition of these representation, due to the default rounding mode, results in a value which differs only in the least-significant-bit.

TL;DR

Writing `0.1 + 0.2` in a IEEE 754 binary representation (with colons separating the three parts) and comparing it to `0.3`, this is (I've put the distinct bits in square brackets):

``````0.1 + 0.2 => 0:01111111101:0011001100110011001100110011001100110011001100110
0.3       => 0:01111111101:0011001100110011001100110011001100110011001100110
``````

Converted back to decimal, these values are:

``````0.1 + 0.2 => 0.300000000000000044408920985006...
0.3       => 0.299999999999999988897769753748...
``````

The difference is exactly 2-54, which is ~5.5511151231258 × 10-17 - insignificant (for many applications) when compared to the original values.

Comparing the last few bits of a floating point number is inherently dangerous, as anyone who reads the famous "What Every Computer Scientist Should Know About Floating-Point Arithmetic" (which covers all the major parts of this answer) will know.

Most calculators use additional guard digits to get around this problem, which is how `0.1 + 0.2` would give `0.3`: the final few bits are rounded.

#### 回答8：

` 问题在于数字可以精确地以10为底，而不能以2为底。这些数字需要四舍五入到最接近的等值。假设使用非常常见的IEEE 64位浮点格式，则最接近0.1的数字是3602879701896397x2⁻⁵⁵，最接近数字的是0.2是7205759403792794x2⁻⁵⁵；将它们加在一起将得到10808639105689191x2⁻⁵⁵，或者精确的十进制值0.3000000000000000444089209850062616169452667236328125。浮点数通常会四舍五入以显示。`

#### A8:

Floating point numbers stored in the computer consist of two parts, an integer and an exponent that the base is taken to and multiplied by the integer part.

If the computer were working in base 10, `0.1` would be `1 x 10⁻¹`, `0.2` would be `2 x 10⁻¹`, and `0.3` would be `3 x 10⁻¹`. Integer math is easy and exact, so adding `0.1 + 0.2` will obviously result in `0.3`.

Computers don't usually work in base 10, they work in base 2. You can still get exact results for some values, for example `0.5` is `1 x 2⁻¹` and `0.25` is `1 x 2⁻²`, and adding them results in `3 x 2⁻²`, or `0.75`. Exactly.

The problem comes with numbers that can be represented exactly in base 10, but not in base 2. Those numbers need to be rounded to their closest equivalent. Assuming the very common IEEE 64-bit floating point format, the closest number to `0.1` is `3602879701896397 x 2⁻⁵⁵`, and the closest number to `0.2` is `7205759403792794 x 2⁻⁵⁵`; adding them together results in `10808639105689191 x 2⁻⁵⁵`, or an exact decimal value of `0.3000000000000000444089209850062616169452667236328125`. Floating point numbers are generally rounded for display.

#### A9:

Floating point rounding error. From What Every Computer Scientist Should Know About Floating-Point Arithmetic:

Squeezing infinitely many real numbers into a finite number of bits requires an approximate representation. Although there are infinitely many integers, in most programs the result of integer computations can be stored in 32 bits. In contrast, given any fixed number of bits, most calculations with real numbers will produce quantities that cannot be exactly represented using that many bits. Therefore the result of a floating-point calculation must often be rounded in order to fit back into its finite representation. This rounding error is the characteristic feature of floating-point computation.

#### 回答10：

``````function add(a, b, precision) {
var x = Math.pow(10, precision || 2);
return (Math.round(a * x) + Math.round(b * x)) / x;
}
``````

#### A10:

My workaround:

``````function add(a, b, precision) {
var x = Math.pow(10, precision || 2);
return (Math.round(a * x) + Math.round(b * x)) / x;
}
``````

precision refers to the number of digits you want to preserve after the decimal point during addition.

#### A11:

A lot of good answers have been posted, but I'd like to append one more.

Not all numbers can be represented via floats/doubles For example, the number "0.2" will be represented as "0.200000003" in single precision in IEEE754 float point standard.

Model for store real numbers under the hood represent float numbers as

Even though you can type `0.2` easily, `FLT_RADIX` and `DBL_RADIX` is 2; not 10 for a computer with FPU which uses "IEEE Standard for Binary Floating-Point Arithmetic (ISO/IEEE Std 754-1985)".

So it is a bit hard to represent such numbers exactly. Even if you specify this variable explicitly without any intermediate calculation.

#### 回答12：

``````0.1 + 0.2 = 0.30000000000000004 (BIGGER)
0.1 + 0.7 = 0.7999999999999999 (SMALLER)
...
1.7 + 1.9 = 3.5999999999999996 (SMALLER)
1.7 + 2.2 = 3.9000000000000004 (BIGGER)
...
3.2 + 3.6 = 6.800000000000001 (BIGGER)
3.2 + 4.4 = 7.6000000000000005 (BIGGER)
``````

``````0.6 - 0.2 = 0.39999999999999997 (SMALLER)
0.5 - 0.4 = 0.09999999999999998 (SMALLER)
...
2.1 - 0.2 = 1.9000000000000001 (BIGGER)
2.0 - 1.9 = 0.10000000000000009 (BIGGER)
...
100 - 99.9 = 0.09999999999999432 (SMALLER)
100 - 99.8 = 0.20000000000000284 (BIGGER)
``````

* 15％和34％的确很大，因此在精度至关重要时，请始终使用BigDecimal。如果使用两位小数(第0.01步)，情况会进一步恶化(分别为18％和36％)。

#### A12:

Some statistics related to this famous double precision question.

When adding all values (a + b) using a step of 0.1 (from 0.1 to 100) we have ~15% chance of precision error. Note that the error could result in slightly bigger or smaller values. Here are some examples:

``````0.1 + 0.2 = 0.30000000000000004 (BIGGER)
0.1 + 0.7 = 0.7999999999999999 (SMALLER)
...
1.7 + 1.9 = 3.5999999999999996 (SMALLER)
1.7 + 2.2 = 3.9000000000000004 (BIGGER)
...
3.2 + 3.6 = 6.800000000000001 (BIGGER)
3.2 + 4.4 = 7.6000000000000005 (BIGGER)
``````

When subtracting all values (a - b where a > b) using a step of 0.1 (from 100 to 0.1) we have ~34% chance of precision error. Here are some examples:

``````0.6 - 0.2 = 0.39999999999999997 (SMALLER)
0.5 - 0.4 = 0.09999999999999998 (SMALLER)
...
2.1 - 0.2 = 1.9000000000000001 (BIGGER)
2.0 - 1.9 = 0.10000000000000009 (BIGGER)
...
100 - 99.9 = 0.09999999999999432 (SMALLER)
100 - 99.8 = 0.20000000000000284 (BIGGER)
``````

*15% and 34% are indeed huge, so always use BigDecimal when precision is of big importance. With 2 decimal digits (step 0.01) the situation worsens a bit more (18% and 36%).

a /(2 n x 5 m )

a / 2 n

### No, not broken, but most decimal fractions must be approximated

Summary

Floating point arithmetic is exact, unfortunately, it doesn't match up well with our usual base-10 number representation, so it turns out we are often giving it input that is slightly off from what we wrote.

Even simple numbers like 0.01, 0.02, 0.03, 0.04 ... 0.24 are not representable exactly as binary fractions. If you count up 0.01, .02, .03 ..., not until you get to 0.25 will you get the first fraction representable in base2. If you tried that using FP, your 0.01 would have been slightly off, so the only way to add 25 of them up to a nice exact 0.25 would have required a long chain of causality involving guard bits and rounding. It's hard to predict so we throw up our hands and say "FP is inexact", but that's not really true.

We constantly give the FP hardware something that seems simple in base 10 but is a repeating fraction in base 2.

How did this happen?

When we write in decimal, every fraction (specifically, every terminating decimal) is a rational number of the form

a / (2n x 5m)

In binary, we only get the 2n term, that is:

a / 2n

So in decimal, we can't represent 1/3. Because base 10 includes 2 as a prime factor, every number we can write as a binary fraction also can be written as a base 10 fraction. However, hardly anything we write as a base10 fraction is representable in binary. In the range from 0.01, 0.02, 0.03 ... 0.99, only three numbers can be represented in our FP format: 0.25, 0.50, and 0.75, because they are 1/4, 1/2, and 3/4, all numbers with a prime factor using only the 2n term.

In base10 we can't represent 1/3. But in binary, we can't do 1/10 or 1/3.

So while every binary fraction can be written in decimal, the reverse is not true. And in fact most decimal fractions repeat in binary.

Dealing with it

Developers are usually instructed to do < epsilon comparisons, better advice might be to round to integral values (in the C library: round() and roundf(), i.e., stay in the FP format) and then compare. Rounding to a specific decimal fraction length solves most problems with output.

Also, on real number-crunching problems (the problems that FP was invented for on early, frightfully expensive computers) the physical constants of the universe and all other measurements are only known to a relatively small number of significant figures, so the entire problem space was "inexact" anyway. FP "accuracy" isn't a problem in this kind of application.

The whole issue really arises when people try to use FP for bean counting. It does work for that, but only if you stick to integral values, which kind of defeats the point of using it. This is why we have all those decimal fraction software libraries.

I love the Pizza answer by Chris, because it describes the actual problem, not just the usual handwaving about "inaccuracy". If FP were simply "inaccurate", we could fix that and would have done it decades ago. The reason we haven't is because the FP format is compact and fast and it's the best way to crunch a lot of numbers. Also, it's a legacy from the space age and arms race and early attempts to solve big problems with very slow computers using small memory systems. (Sometimes, individual magnetic cores for 1-bit storage, but that's another story.)

Conclusion

If you are just counting beans at a bank, software solutions that use decimal string representations in the first place work perfectly well. But you can't do quantum chromodynamics or aerodynamics that way.

#### 回答14：

`````` if( (n * 0.1) < 100.0 ) { return n * 0.1 - 0.000000000000001 ;}
else { return n * 0.1 + 0.000000000000001 ;}
``````

#### A14:

Did you try the duct tape solution?

Try to determine when errors occur and fix them with short if statements, it's not pretty but for some problems it is the only solution and this is one of them.

`````` if( (n * 0.1) < 100.0 ) { return n * 0.1 - 0.000000000000001 ;}
else { return n * 0.1 + 0.000000000000001 ;}
``````

I had the same problem in a scientific simulation project in c#, and I can tell you that if you ignore the butterfly effect it's gonna turn to a big fat dragon and bite you in the a**

#### 回答15：

``````parseFloat((0.1 + 0.2).toFixed(10)) => Will return 0.3
``````

```````0.22 + 0.7 = 0.9199999999999999`
``````

``````(0.2 + 0.7).toFixed(10) => Result will be "0.9000000000"
``````

``````parseFloat((0.2 + 0.7).toFixed(10)) => Result will be 0.9
``````

``````function floatify(number){
return parseFloat((number).toFixed(10));
}
``````

``````function floatify(number){
return parseFloat((number).toFixed(10));
}

var number1 = +\$("#number1").val();
var number2 = +\$("#number2").val();
var unexpectedResult = number1 + number2;
var expectedResult = floatify(number1 + number2);
\$("#unexpectedResult").text(unexpectedResult);
\$("#expectedResult").text(expectedResult);
}

``````input{
width: 50px;
}
#expectedResult{
color: green;
}
#unexpectedResult{
color: red;
}``````

``````<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<p>Expected Result: <span id="expectedResult"></span></p>
<p>Unexpected Result: <span id="unexpectedResult"></span></p>``````

#### A15:

In order to offer The best solution I can say I discovered following method:

``````parseFloat((0.1 + 0.2).toFixed(10)) => Will return 0.3
``````

Let me explain why it's the best solution. As others mentioned in above answers it's a good idea to use ready to use Javascript toFixed() function to solve the problem. But most likely you'll encounter with some problems.

Imagine you are going to add up two float numbers like `0.2` and `0.7` here it is: `0.2 + 0.7 = 0.8999999999999999`.

Your expected result was `0.9` it means you need a result with 1 digit precision in this case. So you should have used `(0.2 + 0.7).tofixed(1)` but you can't just give a certain parameter to toFixed() since it depends on the given number, for instance

```````0.22 + 0.7 = 0.9199999999999999`
``````

In this example you need 2 digits precision so it should be `toFixed(2)`, so what should be the paramter to fit every given float number?

You might say let it be 10 in every situation then:

``````(0.2 + 0.7).toFixed(10) => Result will be "0.9000000000"
``````

Damn! What are you going to do with those unwanted zeros after 9? It's the time to convert it to float to make it as you desire:

``````parseFloat((0.2 + 0.7).toFixed(10)) => Result will be 0.9
``````

Now that you found the solution, it's better to offer it as a function like this:

``````function floatify(number){
return parseFloat((number).toFixed(10));
}
``````

Let's try it yourself:

``````function floatify(number){
return parseFloat((number).toFixed(10));
}

var number1 = +\$("#number1").val();
var number2 = +\$("#number2").val();
var unexpectedResult = number1 + number2;
var expectedResult = floatify(number1 + number2);
\$("#unexpectedResult").text(unexpectedResult);
\$("#expectedResult").text(expectedResult);
}
``````input{
width: 50px;
}
#expectedResult{
color: green;
}
#unexpectedResult{
color: red;
}``````
``````<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<p>Expected Result: <span id="expectedResult"></span></p>
<p>Unexpected Result: <span id="unexpectedResult"></span></p>``````

You can use it this way:

``````var x = 0.2 + 0.7;
floatify(x);  => Result: 0.9
``````

As W3SCHOOLS suggests there is another solution too, you can multiply and divide to solve the problem above:

``````var x = (0.2 * 10 + 0.1 * 10) / 10;       // x will be 0.3
``````

Keep in mind that `(0.2 + 0.1) * 10 / 10` won't work at all although it seems the same! I prefer the first solution since I can apply it as a function which converts the input float to accurate output float.

#### A16:

Those weird numbers appear because computers use binary(base 2) number system for calculation purposes, while we use decimal(base 10).

There are a majority of fractional numbers that cannot be represented precisely either in binary or in decimal or both. Result - A rounded up (but precise) number results.

#### 回答17：

0.1转换为0.1000000000000000055511151231257827021181583404541015625，

0.2转换为0.200000000000000011102230246251565404236316680908203125，

0.3转换为0.299999999999999988897769753748434595763683319091796875和

0.30000000000000004转换为0.3000000000000000444089209850062616169452667236328125。

#### A17:

Many of this question's numerous duplicates ask about the effects of floating point rounding on specific numbers. In practice, it is easier to get a feeling for how it works by looking at exact results of calculations of interest rather than by just reading about it. Some languages provide ways of doing that - such as converting a `float` or `double` to `BigDecimal` in Java.

Since this is a language-agnostic question, it needs language-agnostic tools, such as a Decimal to Floating-Point Converter.

Applying it to the numbers in the question, treated as doubles:

0.1 converts to 0.1000000000000000055511151231257827021181583404541015625,

0.2 converts to 0.200000000000000011102230246251565404236316680908203125,

0.3 converts to 0.299999999999999988897769753748434595763683319091796875, and

0.30000000000000004 converts to 0.3000000000000000444089209850062616169452667236328125.

Adding the first two numbers manually or in a decimal calculator such as Full Precision Calculator, shows the exact sum of the actual inputs is 0.3000000000000000166533453693773481063544750213623046875.

If it were rounded down to the equivalent of 0.3 the rounding error would be 0.0000000000000000277555756156289135105907917022705078125. Rounding up to the equivalent of 0.30000000000000004 also gives rounding error 0.0000000000000000277555756156289135105907917022705078125. The round-to-even tie breaker applies.

Returning to the floating point converter, the raw hexadecimal for 0.30000000000000004 is 3fd3333333333334, which ends in an even digit and therefore is the correct result.

#### 回答18：

• Python的 `十进制`模块和Java的 `BigDecimal`，内部以十进制表示法表示数字(与二进制表示法相对)。两者的精度都有限，因此它们仍然容易出错，但是它们可以解决二进制浮点算术中最常见的问题。

在处理货币时，小数非常好：十美分加二十美分始终恰好是三十美分：

``````>>> 0.1 + 0.2 == 0.3
False
>>> Decimal('0.1') + Decimal('0.2') == Decimal('0.3')
True
``````

Python的`decimal`模块基于 IEEE标准854-1987

• Python的 `fractions`模块和Apache Common的 ```BigFraction 类````。两者都将有理数表示为(分子,分母)对，并且它们给出的结果可能比十进制浮点算术更准确。`

` `
• ` `
` 这两种解决方案都不是完美的(特别是如果我们考虑性能，或者需要非常高的精度)，但是它们仍然使用二进制浮点算法解决了许多问题。`

#### A18:

Given that nobody has mentioned this...

Some high level languages such as Python and Java come with tools to overcome binary floating point limitations. For example:

• Python's `decimal` module and Java's `BigDecimal` class, that represent numbers internally with decimal notation (as opposed to binary notation). Both have limited precision, so they are still error prone, however they solve most common problems with binary floating point arithmetic.

Decimals are very nice when dealing with money: ten cents plus twenty cents are always exactly thirty cents:

``````>>> 0.1 + 0.2 == 0.3
False
>>> Decimal('0.1') + Decimal('0.2') == Decimal('0.3')
True
``````

Python's `decimal` module is based on IEEE standard 854-1987.

• Python's `fractions` module and Apache Common's `BigFraction` class. Both represent rational numbers as `(numerator, denominator)` pairs and they may give more accurate results than decimal floating point arithmetic.

Neither of these solutions is perfect (especially if we look at performances, or if we require a very high precision), but still they solve a great number of problems with binary floating point arithmetic.

#### A19:

Can I just add; people always assume this to be a computer problem, but if you count with your hands (base 10), you can't get `(1/3+1/3=2/3)=true` unless you have infinity to add 0.333... to 0.333... so just as with the `(1/10+2/10)!==3/10` problem in base 2, you truncate it to 0.333 + 0.333 = 0.666 and probably round it to 0.667 which would be also be technically inaccurate.

Count in ternary, and thirds are not a problem though - maybe some race with 15 fingers on each hand would ask why your decimal math was broken...

#### A20:

The kind of floating-point math that can be implemented in a digital computer necessarily uses an approximation of the real numbers and operations on them. (The standard version runs to over fifty pages of documentation and has a committee to deal with its errata and further refinement.)

This approximation is a mixture of approximations of different kinds, each of which can either be ignored or carefully accounted for due to its specific manner of deviation from exactitude. It also involves a number of explicit exceptional cases at both the hardware and software levels that most people walk right past while pretending not to notice.

If you need infinite precision (using the number π, for example, instead of one of its many shorter stand-ins), you should write or use a symbolic math program instead.

But if you're okay with the idea that sometimes floating-point math is fuzzy in value and logic and errors can accumulate quickly, and you can write your requirements and tests to allow for that, then your code can frequently get by with what's in your FPU.

#### 回答21：

``````SIGN EXPONENT FRACTION
``````

``````#include <stdio.h>
#include <limits.h>

void
xx(float *x)
{
unsigned char i = sizeof(*x)*CHAR_BIT-1;
do {
switch (i) {
case 31:
printf("sign:");
break;
case 30:
printf("exponent:");
break;
case 23:
printf("fraction:");
break;

}
char b=(*(unsigned long long*)x&((unsigned long long)1<<i))!=0;
printf("%d ", b);
} while (i--);
printf("\n");
}

void
yy(float a)
{
int sign=!(*(unsigned long long*)&a&((unsigned long long)1<<31));
int fraction = ((1<<23)-1)&(*(int*)&a);
int exponent = (255&((*(int*)&a)>>23))-127;

printf(sign?"positive" " ( 1+":"negative" " ( 1+");
unsigned int i = 1<<22;
unsigned int j = 1;
do {
char b=(fraction&i)!=0;
b&&(printf("1/(%d) %c", 1<<j, (fraction&(i-1))?'+':')' ), 0);
} while (j++, i>>=1);

printf("*2^%d", exponent);
printf("\n");
}

void
main()
{
float x=-3.14;
float y=999999999;
printf("%lu\n", sizeof(x));
xx(&x);
xx(&y);
yy(x);
yy(y);
}
``````

``````-- .../terra1/stub
@ qemacs f.c
-- .../terra1/stub
@ gcc f.c
-- .../terra1/stub
@ ./a.out
sign:1 exponent:1 0 0 0 0 0 0 fraction:0 1 0 0 1 0 0 0 1 1 1 1 0 1 0 1 1 1 0 0 0 0 1 1
sign:0 exponent:1 0 0 1 1 1 0 fraction:0 1 1 0 1 1 1 0 0 1 1 0 1 0 1 1 0 0 1 0 1 0 0 0
negative ( 1+1/(2) +1/(16) +1/(256) +1/(512) +1/(1024) +1/(2048) +1/(8192) +1/(32768) +1/(65536) +1/(131072) +1/(4194304) +1/(8388608) )*2^1
positive ( 1+1/(2) +1/(4) +1/(16) +1/(32) +1/(64) +1/(512) +1/(1024) +1/(4096) +1/(16384) +1/(32768) +1/(262144) +1/(1048576) )*2^29
-- .../terra1/stub
@ bc
scale=15
( 1+1/(2) +1/(4) +1/(16) +1/(32) +1/(64) +1/(512) +1/(1024) +1/(4096) +1/(16384) +1/(32768) +1/(262144) +1/(1048576) )*2^29
999999999.999999446351872
``````

``````999999999.999999446351872
``````

#### A21:

Just for fun, I played with the representation of floats, following the definitions from the Standard C99 and I wrote the code below.

The code prints the binary representation of floats in 3 separated groups

``````SIGN EXPONENT FRACTION
``````

and after that it prints a sum, that, when summed with enough precision, it will show the value that really exists in hardware.

So when you write `float x = 999...`, the compiler will transform that number in a bit representation printed by the function `xx` such that the sum printed by the function `yy` be equal to the given number.

In reality, this sum is only an approximation. For the number 999,999,999 the compiler will insert in bit representation of the float the number 1,000,000,000

After the code I attach a console session, in which I compute the sum of terms for both constants (minus PI and 999999999) that really exists in hardware, inserted there by the compiler.

``````#include <stdio.h>
#include <limits.h>

void
xx(float *x)
{
unsigned char i = sizeof(*x)*CHAR_BIT-1;
do {
switch (i) {
case 31:
printf("sign:");
break;
case 30:
printf("exponent:");
break;
case 23:
printf("fraction:");
break;

}
char b=(*(unsigned long long*)x&((unsigned long long)1<<i))!=0;
printf("%d ", b);
} while (i--);
printf("\n");
}

void
yy(float a)
{
int sign=!(*(unsigned long long*)&a&((unsigned long long)1<<31));
int fraction = ((1<<23)-1)&(*(int*)&a);
int exponent = (255&((*(int*)&a)>>23))-127;

printf(sign?"positive" " ( 1+":"negative" " ( 1+");
unsigned int i = 1<<22;
unsigned int j = 1;
do {
char b=(fraction&i)!=0;
b&&(printf("1/(%d) %c", 1<<j, (fraction&(i-1))?'+':')' ), 0);
} while (j++, i>>=1);

printf("*2^%d", exponent);
printf("\n");
}

void
main()
{
float x=-3.14;
float y=999999999;
printf("%lu\n", sizeof(x));
xx(&x);
xx(&y);
yy(x);
yy(y);
}
``````

Here is a console session in which I compute the real value of the float that exists in hardware. I used `bc` to print the sum of terms outputted by the main program. One can insert that sum in python `repl` or something similar also.

``````-- .../terra1/stub
@ qemacs f.c
-- .../terra1/stub
@ gcc f.c
-- .../terra1/stub
@ ./a.out
sign:1 exponent:1 0 0 0 0 0 0 fraction:0 1 0 0 1 0 0 0 1 1 1 1 0 1 0 1 1 1 0 0 0 0 1 1
sign:0 exponent:1 0 0 1 1 1 0 fraction:0 1 1 0 1 1 1 0 0 1 1 0 1 0 1 1 0 0 1 0 1 0 0 0
negative ( 1+1/(2) +1/(16) +1/(256) +1/(512) +1/(1024) +1/(2048) +1/(8192) +1/(32768) +1/(65536) +1/(131072) +1/(4194304) +1/(8388608) )*2^1
positive ( 1+1/(2) +1/(4) +1/(16) +1/(32) +1/(64) +1/(512) +1/(1024) +1/(4096) +1/(16384) +1/(32768) +1/(262144) +1/(1048576) )*2^29
-- .../terra1/stub
@ bc
scale=15
( 1+1/(2) +1/(4) +1/(16) +1/(32) +1/(64) +1/(512) +1/(1024) +1/(4096) +1/(16384) +1/(32768) +1/(262144) +1/(1048576) )*2^29
999999999.999999446351872
``````

That's it. The value of 999999999 is in fact

``````999999999.999999446351872
``````

You can also check with `bc` that -3.14 is also perturbed. Do not forget to set a `scale` factor in `bc`.

The displayed sum is what inside the hardware. The value you obtain by computing it depends on the scale you set. I did set the `scale` factor to 15. Mathematically, with infinite precision, it seems it is 1,000,000,000.

#### A22:

Another way to look at this: Used are 64 bits to represent numbers. As consequence there is no way more than 2**64 = 18,446,744,073,709,551,616 different numbers can be precisely represented.

However, Math says there are already infinitely many decimals between 0 and 1. IEE 754 defines an encoding to use these 64 bits efficiently for a much larger number space plus NaN and +/- Infinity, so there are gaps between accurately represented numbers filled with numbers only approximated.

Unfortunately 0.3 sits in a gap.

#### A23:

Since this thread branched off a bit into a general discussion over current floating point implementations I'd add that there are projects on fixing their issues.

Take a look at https://posithub.org/ for example, which showcases a number type called posit (and its predecessor unum) that promises to offer better accuracy with fewer bits. If my understanding is correct, it also fixes the kind of problems in the question. Quite interesting project, the person behind it is a mathematician it Dr. John Gustafson. The whole thing is open source, with many actual implementations in C/C++, Python, Julia and C# (https://hastlayer.com/arithmetics).

#### 回答24：

``````1/3 + 2 / 3 == 1
``````

1/3 = 0.333 .... 2/3 = 0.666 ....

``````0.33333333 + 0.66666666 = 0.99999999
``````

1/10 = 0.0001100110011001100 ...(以2为底)

1/5 = 0.0011001100110011001 ...(以2为底)

``````0.0001100 + 0.0011001 = 0.0100101
``````

3/10 = 0.01001100110011 ...(以2为底)

`0.0100110`，它们之间的差异恰好是`0.0000001`

#### A24:

Imagine working in base ten with, say, 8 digits of accuracy. You check whether

``````1/3 + 2 / 3 == 1
``````

and learn that this returns `false`. Why? Well, as real numbers we have

1/3 = 0.333.... and 2/3 = 0.666....

Truncating at eight decimal places, we get

``````0.33333333 + 0.66666666 = 0.99999999
``````

which is, of course, different from `1.00000000` by exactly `0.00000001`.

The situation for binary numbers with a fixed number of bits is exactly analogous. As real numbers, we have

1/10 = 0.0001100110011001100... (base 2)

and

1/5 = 0.0011001100110011001... (base 2)

If we truncated these to, say, seven bits, then we'd get

``````0.0001100 + 0.0011001 = 0.0100101
``````

while on the other hand,

3/10 = 0.01001100110011... (base 2)

which, truncated to seven bits, is `0.0100110`, and these differ by exactly `0.0000001`.

The exact situation is slightly more subtle because these numbers are typically stored in scientific notation. So, for instance, instead of storing 1/10 as `0.0001100` we may store it as something like `1.10011 * 2^-4`, depending on how many bits we've allocated for the exponent and the mantissa. This affects how many digits of precision you get for your calculations.

The upshot is that because of these rounding errors you essentially never want to use == on floating-point numbers. Instead, you can check if the absolute value of their difference is smaller than some fixed small number.

#### A25:

It's actually pretty simple. When you have a base 10 system (like ours), it can only express fractions that use a prime factor of the base. The prime factors of 10 are 2 and 5. So 1/2, 1/4, 1/5, 1/8, and 1/10 can all be expressed cleanly because the denominators all use prime factors of 10. In contrast, 1/3, 1/6, and 1/7 are all repeating decimals because their denominators use a prime factor of 3 or 7. In binary (or base 2), the only prime factor is 2. So you can only express fractions cleanly which only contain 2 as a prime factor. In binary, 1/2, 1/4, 1/8 would all be expressed cleanly as decimals. While, 1/5 or 1/10 would be repeating decimals. So 0.1 and 0.2 (1/10 and 1/5) while clean decimals in a base 10 system, are repeating decimals in the base 2 system the computer is operating in. When you do math on these repeating decimals, you end up with leftovers which carry over when you convert the computer's base 2 (binary) number into a more human readable base 10 number.

#### 回答26：

``````>>> import math
>>> math.isclose(0.1 + 0.2, 0.3)
True
>>> 0.1 + 0.2 == 0.3
False
``````

#### A26:

Since Python 3.5 you can use `math.isclose()` function for testing approximate equality:

``````>>> import math
>>> math.isclose(0.1 + 0.2, 0.3)
True
>>> 0.1 + 0.2 == 0.3
False
``````

#### 回答27：

``````#include <stdio.h>

int main() {
printf("0.1 + 0.2 == 0.3 is %s\n", 0.1 + 0.2 == 0.3 ? "true" : "false");
printf("0.1 is %.23f\n", 0.1);
printf("0.2 is %.23f\n", 0.2);
printf("0.1 + 0.2 is %.23f\n", 0.1 + 0.2);
printf("0.3 is %.23f\n", 0.3);
printf("0.3 - (0.1 + 0.2) is %g\n", 0.3 - (0.1 + 0.2));
return 0;
}
``````

``````0.1 + 0.2 == 0.3 is false
0.1 is 0.10000000000000000555112
0.2 is 0.20000000000000001110223
0.1 + 0.2 is 0.30000000000000004440892
0.3 is 0.29999999999999998889777
0.3 - (0.1 + 0.2) is -5.55112e-17
``````

`_Decimal32``_Decimal64``_Decimal128`类型可能在您的系统上可用(例如， GCC 选定的目标，但 lang语 OS X )上不支持它们。

#### A27:

Decimal numbers such as `0.1`, `0.2`, and `0.3` are not represented exactly in binary encoded floating point types. The sum of the approximations for `0.1` and `0.2` differs from the approximation used for `0.3`, hence the falsehood of `0.1 + 0.2 == 0.3` as can be seen more clearly here:

``````#include <stdio.h>

int main() {
printf("0.1 + 0.2 == 0.3 is %s\n", 0.1 + 0.2 == 0.3 ? "true" : "false");
printf("0.1 is %.23f\n", 0.1);
printf("0.2 is %.23f\n", 0.2);
printf("0.1 + 0.2 is %.23f\n", 0.1 + 0.2);
printf("0.3 is %.23f\n", 0.3);
printf("0.3 - (0.1 + 0.2) is %g\n", 0.3 - (0.1 + 0.2));
return 0;
}
``````

Output:

``````0.1 + 0.2 == 0.3 is false
0.1 is 0.10000000000000000555112
0.2 is 0.20000000000000001110223
0.1 + 0.2 is 0.30000000000000004440892
0.3 is 0.29999999999999998889777
0.3 - (0.1 + 0.2) is -5.55112e-17
``````

For these computations to be evaluated more reliably, you would need to use a decimal-based representation for floating point values. The C Standard does not specify such types by default but as an extension described in a technical Report.

The `_Decimal32`, `_Decimal64` and `_Decimal128` types might be available on your system (for example, GCC supports them on selected targets, but Clang does not support them on OSX).

#### 回答28：

Math.sum (javascript)....一种替换运算符

``````.1 + .0001 + -.1 --> 0.00010000000000000286
Math.sum(.1 , .0001, -.1) --> 0.0001
``````

``````Object.defineProperties(Math, {
sign: {
value: function (x) {
return x ? x < 0 ? -1 : 1 : 0;
}
},
precision: {
value: function (value, precision, type) {
var v = parseFloat(value),
p = Math.max(precision, 0) || 0,
t = type || 'round';
return (Math[t](v * Math.pow(10, p)) / Math.pow(10, p)).toFixed(p);
}
},
scientific_to_num: {  // this is from https://gist.github.com/jiggzson
value: function (num) {
//if the number is in scientific notation remove it
if (/e/i.test(num)) {
var zero = '0',
parts = String(num).toLowerCase().split('e'), //split into coeff and exponent
e = parts.pop(), //store the exponential part
l = Math.abs(e), //get the number of zeros
sign = e / l,
coeff_array = parts.split('.');
if (sign === -1) {
num = zero + '.' + new Array(l).join(zero) + coeff_array.join('');
} else {
var dec = coeff_array;
if (dec)
l = l - dec.length;
num = coeff_array.join('') + new Array(l + 1).join(zero);
}
}
return num;
}
}
get_precision: {
value: function (number) {
var arr = Math.scientific_to_num((number + "")).split(".");
return arr ? arr.length : 0;
}
},
diff:{
value: function(A,B){
var prec = this.max(this.get_precision(A),this.get_precision(B));
return +this.precision(A-B,prec);
}
},
sum: {
value: function () {
var prec = 0, sum = 0;
for (var i = 0; i < arguments.length; i++) {
prec = this.max(prec, this.get_precision(arguments[i]));
sum += +arguments[i]; // force float to convert strings to number
}
return Math.precision(sum, prec);
}
}
});
``````

``````Math.diff(0.2, 0.11) == 0.09 // true
0.2 - 0.11 == 0.09 // false
``````

Math.sum接受任意数量的参数

#### A28:

Math.sum ( javascript ) .... kind of operator replacement

``````.1 + .0001 + -.1 --> 0.00010000000000000286
Math.sum(.1 , .0001, -.1) --> 0.0001
``````

``````Object.defineProperties(Math, {
sign: {
value: function (x) {
return x ? x < 0 ? -1 : 1 : 0;
}
},
precision: {
value: function (value, precision, type) {
var v = parseFloat(value),
p = Math.max(precision, 0) || 0,
t = type || 'round';
return (Math[t](v * Math.pow(10, p)) / Math.pow(10, p)).toFixed(p);
}
},
scientific_to_num: {  // this is from https://gist.github.com/jiggzson
value: function (num) {
//if the number is in scientific notation remove it
if (/e/i.test(num)) {
var zero = '0',
parts = String(num).toLowerCase().split('e'), //split into coeff and exponent
e = parts.pop(), //store the exponential part
l = Math.abs(e), //get the number of zeros
sign = e / l,
coeff_array = parts.split('.');
if (sign === -1) {
num = zero + '.' + new Array(l).join(zero) + coeff_array.join('');
} else {
var dec = coeff_array;
if (dec)
l = l - dec.length;
num = coeff_array.join('') + new Array(l + 1).join(zero);
}
}
return num;
}
}
get_precision: {
value: function (number) {
var arr = Math.scientific_to_num((number + "")).split(".");
return arr ? arr.length : 0;
}
},
diff:{
value: function(A,B){
var prec = this.max(this.get_precision(A),this.get_precision(B));
return +this.precision(A-B,prec);
}
},
sum: {
value: function () {
var prec = 0, sum = 0;
for (var i = 0; i < arguments.length; i++) {
prec = this.max(prec, this.get_precision(arguments[i]));
sum += +arguments[i]; // force float to convert strings to number
}
return Math.precision(sum, prec);
}
}
});
``````

the idea is to use Math instead operators to avoid float errors

``````Math.diff(0.2, 0.11) == 0.09 // true
0.2 - 0.11 == 0.09 // false
``````

also note that Math.diff and Math.sum auto-detect the precision to use

Math.sum accepts any number of arguments

# 我刚刚看到了有关浮点数的有趣问题：

``````error = (2**53+1) - int(float(2**53+1))
``````
``````>>> (2**53+1) - int(float(2**53+1))
1
``````

`2**53+1`时，我们可以清楚地看到一个断点-直到`2**53`都可以正常工作。

``````>>> (2**53) - int(float(2**53))
0
``````

• 符号位：1位
• 指数：11位
• 极高的精度：53位(显式存储52位)

# I just saw this interesting issue around floating points:

Consider the following results:

``````error = (2**53+1) - int(float(2**53+1))
``````
``````>>> (2**53+1) - int(float(2**53+1))
1
``````

We can clearly see a breakpoint when `2**53+1` - all works fine until `2**53`.

``````>>> (2**53) - int(float(2**53))
0
``````

This happens because of the double-precision binary: IEEE 754 double-precision binary floating-point format: binary64

Double-precision binary floating-point is a commonly used format on PCs, due to its wider range over single-precision floating point, in spite of its performance and bandwidth cost. As with single-precision floating-point format, it lacks precision on integer numbers when compared with an integer format of the same size. It is commonly known simply as double. The IEEE 754 standard specifies a binary64 as having:

• Sign bit: 1 bit
• Exponent: 11 bits
• Significant precision: 53 bits (52 explicitly stored)

The real value assumed by a given 64-bit double-precision datum with a given biased exponent and a 52-bit fraction is

or

Thanks to @a_guest for pointing that out to me.

#### 回答30：

``````float x = 9.9F;
``````

#### A30:

A different question has been named as a duplicate to this one:

In C++, why is the result of `cout << x` different from the value that a debugger is showing for `x`?

The `x` in the question is a `float` variable.

One example would be

``````float x = 9.9F;
``````

The debugger shows `9.89999962`, the output of `cout` operation is `9.9`.

The answer turns out to be that `cout`'s default precision for `float` is 6, so it rounds to 6 decimal digits.

See here for reference