由于float和double的精度有限，导致错误 [英] Error due to limited precision of float and double

问题描述

在C ++中，由于float和double的精度有限，我使用以下代码计算错误的数量级：

  float n = 1; 
 float dec = 1; 
 
 while（n！=（n-dec））{
 dec = dec / 10; 
} 
 cout<< dec<< endl; 
  

（在双重情况下，我所做的是在第1行和第2行交换float double） p>

现在，当我在Unix系统上使用g ++编译和运行它时，结果是

  Float 10 ^ -8 
 Double 10 ^ -17 
  

我在Windows 7上使用MinGW编译和运行它，结果是

 浮点数10 ^ -20 
双10 ^ -20 
  

这是什么原因？

解决方案

我想我会让我的评论一个答案，并扩大它。 这是我的假设，我可能会错。

Windows上的MinGW可能通过将表达式的中间体因此，表达式 n！=（n-dec）的两边都可以使用x86的80位精度。

<被评估为64位精度（80位FP具有64位尾数）。

  2 ^ -64〜 10 ^ -20 
  

所以数字是有意义的。

Visual Studio（默认情况下）也会促进中间体。但只能达到双精度。

In C++, I use the following code to work out the order of magnitude of the error due to the limited precision of float and double:

 float n=1;
 float dec  = 1;

 while(n!=(n-dec)) {
    dec = dec/10;
 }
 cout << dec << endl;

(in the double case all I do is exchange float with double in line 1 and 2)

Now when I compile and run this using g++ on a Unix system, the results are

Float  10^-8
Double 10^-17

However, when I compile and run it using MinGW on Windows 7, the results are

Float  10^-20
Double 10^-20

What is the reason for this?

解决方案

I guess I'll make my comment an answer and expand on it. This is my hypothesis, I may be wrong.

MinGW on Windows is probably trying to preserve precision by promoting the intermediates of expressions to the full 80-bit precision of x86.

Therefore, both sides of the expression n != (n-dec) are evaluated to 64-bits of precision (80-bit FP has a 64-bit mantissa).

2^-64 ~ 10^-20

So the numbers make sense.

Visual Studio also (by default), will promote intermediates. But only up to double-precision.

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