浮点比较 `a != 0.7` [英] Floating point comparison `a != 0.7`

问题描述

可能重复:
浮点比较中的问题

#include <stdio.h>
#include <conio.h>

main()
{
    float a = 0.7;
    if(a < 0.7)
        printf("C");
    else
        printf("C++");
}

在上面的代码中，输出是C.我在 Code::Blocks 和 Pelles C 中尝试了这段代码，但得到了相同的答案.我想详细了解一下原因！

In the above code, the output is C. I tried this code in Code::Blocks and Pelles C but got the same answer. I would like know the reason for this in detail!

推荐答案

在二进制中，0.7 是:

In binary, 0.7 is:

b0.1011001100110011001100110011001100110011001100110011001100110...

但是，0.7 是一个双精度字面量，其值是 0.7 舍入到最接近的可表示双精度值，即:

However, 0.7 is a double-precision literal, whose value is 0.7 rounded to the closest representable double-precision value, which is:

b0.10110011001100110011001100110011001100110011001100110

十进制，就是这样:

 0.6999999999999999555910790149937383830547332763671875

当您编写 float a = 0.7 时，该双精度值再次四舍五入为单精度，而 a 获取二进制值:

When you write float a = 0.7, that double value is rounded again to single-precision, and a gets the binary value:

b0.101100110011001100110011

正是

 0.699999988079071044921875

十进制.

当你进行比较 (a < 0.7) 时，你是在比较这个单精度值(转换为双精度，它不会四舍五入，因为所有单精度值都可以用双精度表示精度)到原来的双精度值.因为

When you do the comparison (a < 0.7), you are comparing this single-precision value (converted to double, which does not round, because all single-precision values are representable in double precision) to the original double-precision value. Because

 0.699999988079071044921875 < 0.6999999999999999555910790149937383830547332763671875

比较正确返回 true，并且您的程序打印 "C".

the comparison correctly returns true, and your program prints "C".

请注意，这在 C++ 中没有任何不同，所讨论的代码的外观恰恰相反.有某些(数值上不安全的)编译器优化可以改变行为，但这些并不是 C 或 C++ 独有的.

Please note that none of this is any different in C++, appearances of the code in question to the contrary. There are certain (numerically unsafe) compiler optimizations that can change the behavior, but those are not unique to C or C++.

这篇关于浮点比较 `a != 0.7`的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！