Haskell：表达式的意外输出[0，0.1 .. 1] [英] Haskell: Unexpected output for expression [0, 0.1 .. 1]

问题描述

当评估表达式时：

  * main> [0，0.1 .. 1] 
  

我实际上期待：

  [0,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1] 
  

但是我很震惊地看到输出是

  [0.0,0.1,0.2,0.30000000000000004,0.4000000000000001,0.5000000000000001,0.6000000000000001,0.7000000000000001,0.8,0.9,1.0] 
  

$ b $为什么Haskell在评估后产生这样的结果？ 这是由于浮点值不精确的结果，Haskell并不特别。如果你不能处理浮点数的近似值，那么你可以使用：

 >导入Data.Ratio 
 Data.Ratio> [0,1％10 .. 1％1] 
 [0％1,1％10,1％5,3％10,2％5,1％2,3％5,7％10,4 ％5,9％10,1％1] 
  

 >>> 0.3 
 0.29999999999999999 
  

这里是C：

  void main（）{printf（％0.17f\\\
，0.3）; } 
 
 $ gcc t.c 2> / dev / null; ./a.out 
 0.29999999999999999 
  

When evaluating the expression:

*main> [0, 0.1 .. 1]

I was actually expecting:

 [0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1]

But I was quite shocked to see the output be

[0.0,0.1,0.2,0.30000000000000004,0.4000000000000001,0.5000000000000001,0.6000000000000001,0.7000000000000001,0.8,0.9,1.0]

Why does Haskell produce that result upon evaluation?

解决方案

This is a result of the imprecision of floating point values, it isn't particular to Haskell. If you can't deal with the approximation inherent in floating point then you can use Rational at a high performance cost:

> import Data.Ratio
Data.Ratio> [0,1%10.. 1%1]
[0 % 1,1 % 10,1 % 5,3 % 10,2 % 5,1 % 2,3 % 5,7 % 10,4 % 5,9 % 10,1 % 1]

Just to hammer the point home, here's Python:

>>> 0.3
0.29999999999999999

And here's C:

void main() { printf("%0.17f\n",0.3); }

$ gcc t.c 2>/dev/null ; ./a.out
0.29999999999999999

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