Version: 3.2.2

# 14. 浮点算术: 问题和限制¶

```0.125
```

```0.001
```

```0.3
```

```0.33
```

```0.333
```

```0.0001100110011001100110011001100110011001100110011...
```

```>>> 0.1
0.1000000000000000055511151231257827021181583404541015625
```

```>>> 1 / 10
0.1
```

```>>> format(math.pi, '.12g')  # give 12 significant digits
'3.14159265359'

>>> format(math.pi, '.2f')   # give 2 digits after the point
'3.14'

>>> repr(math.pi)
'3.141592653589793'
```

```>>> .1 + .1 + .1 == .3
False
```

```>>> round(.1, 1) + round(.1, 1) + round(.1, 1) == round(.3, 1)
False
```

```>>> round(.1 + .1 + .1, 10) == round(.3, 10)
True
```

Python 提供了工具来帮助你获得浮点数的准确值. 你可以使用 `float.as_integer_ratio()` 方法来表示一个分数:

```>>> x = 3.14159
>>> x.as_integer_ratio()
(3537115888337719, 1125899906842624)
```

```>>> x == 3537115888337719 / 1125899906842624
True
```

`float.hex()` 方法以十六进制表述, 这也同样给出了一个被你计算机准确存储的值:

```>>> x.hex()
'0x1.921f9f01b866ep+1'
```

```>>> x == float.fromhex('0x1.921f9f01b866ep+1')
True
```

```>>> sum([0.1] * 10) == 1.0
False
>>> math.fsum([0.1] * 10) == 1.0
True
```

## 14.1. 表示错误¶

Representation error 涉及到这样的事实, 有些 (更准确来书是大多数) 小数的分数表示不能够以二进制为底的分数表述. 这就是主要的原因, 为何 Python (或者 Perl, C, C++, Java, Fortran, 还有更多的) 常常不能够如你所愿的表示.

```1 / 10 ~= J / (2**N)
```

```J ~= 2**N / 10
```

```>>> 2**52 <=  2**56 // 10  < 2**53
True
```

```>>> q, r = divmod(2**56, 10)
>>> r
6
```

```>>> q+1
7205759403792794
```

```7205759403792794 / 2 ** 56
```

```3602879701896397 / 2 ** 55
```

```>>> 0.1 * 2 ** 55
3602879701896397.0
```

```>>> 3602879701896397 * 10 ** 55 // 2 ** 55
1000000000000000055511151231257827021181583404541015625
```

```>>> format(0.1, '.17f')
'0.10000000000000001'
```

`fractions``decimal` 模块使这些计算变得简单:

```>>> from decimal import Decimal
>>> from fractions import Fraction

>>> Fraction.from_float(0.1)
Fraction(3602879701896397, 36028797018963968)

>>> (0.1).as_integer_ratio()
(3602879701896397, 36028797018963968)

>>> Decimal.from_float(0.1)
Decimal('0.1000000000000000055511151231257827021181583404541015625')

>>> format(Decimal.from_float(0.1), '.17')
'0.10000000000000001'
```