比较浮点数和双精度数 [英] Comparing float and double

问题描述

#include <stdio.h>
int main(void){
    float a = 1.1;
    double b = 1.1;
    if(a == b){
        printf("if block");
    }
    else{
        printf("else block");
    }
    return 0;
}

打印:否则阻止

#include <stdio.h>
int main(void){
    float a = 1.5;
    double b = 1.5;
    if(a == b){
        printf("if block");
    }
    else{
        printf("else block");
    }
    return 0;
}

打印:如果阻塞

这背后的逻辑是什么?

使用的编译器:gcc-4.3.4

Compiler used: gcc-4.3.4

推荐答案

这是因为 1.1 不能完全用二进制浮点表示.但是 1.5 是.

This is because 1.1 is not exactly representable in binary floating-point. But 1.5 is.

因此，float 和 double 表示将持有稍有不同的 1.1 值.

As a result, the float and double representations will hold slightly different values of 1.1.

这正是写成二进制浮点数时的区别:

Here is exactly the difference when written out as binary floating-point:

(float) 1.1 = (0.00011001100110011001101)₂
(double)1.1 = (0.0001100110011001100110011001100110011001100110011010)₂

因此，当您比较它们时(并且 float 版本得到提升)，它们将不相等.

Thus, when you compare them (and the float version gets promoted), they will not be equal.

这篇关于比较浮点数和双精度数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！