gcc double printf precision - 错误的输出 [英] gcc double printf precision - wrong output

问题描述

  #include< stdio.h> 
 #include< wchar.h> 
 int main（）
 {
 double f = 1717.1800000000001; 
 wprintf（Ldouble％.20G \\\
，f）; 
返回0; 
 
 
 
 
 $ b $ p 
输出（和预期的下面）：
 
 
  double 1717.1800000000000637 
 double 1717.1800000000001 
  
 
 
 <这是在Ubuntu 11.10 x64上（也是在编译32位时）。
 
 
我试图解决的问题是在Windows上输出数字像在代码中一样，我需要使低级格式化（swprintf）像在Windows中一样工作，以实现可移植性问题。    解决方案

1717.1800000000001并不能完全表示为double，所以在f中你只能得到一个值。存储在f中的值恰好是1717.180000000000063664629124104976654052734375。现在的问题是，Windows只输出17位有效数字，虽然有20位被请求（这是一个已知的错误，AFAIK它的错误数据库中的某处）。



如果您不能将字段长度限制为合理的值（如％。17G），你需要一个包装来模仿这个bug。

#include <stdio.h>
#include <wchar.h> 
int main()
{
    double f = 1717.1800000000001;
    wprintf(L"double      %.20G\n", f);
    return 0;
}

outputs (and expected below):

double      1717.1800000000000637
double      1717.1800000000001

This is on Ubuntu 11.10 x64 (but also when compiling for 32 bit).

The problem that I try to solve, is that on Windows it outputs the number exactly like in code, and I need to make low-level formatting (swprintf) to work like in Windows, for portability issues.

解决方案

1717.1800000000001 is not exactly representable as double, so you only get a value near to it in f. The value stored in f is exactly 1717.180000000000063664629124104976654052734375.

The problem is now that windows does only output 17 significant digits, although 20 were requested (which is a known bug, AFAIK it's somewhere in their bug database).

If you can't limit the field length to a sane value (like "%.17G"), you need a wrapper to mimic this bug.

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