C- 浮点精度 [英] C- Floating point precision

问题描述

我有一个程序:

int main() 
{   
        float f = 0.0f;  
        int i;  

        for (i = 0 ; i < 10 ; i++) 
                f = f + 0.1f; 

        if (f == 1.0f) 
                printf("f is 1.0 
"); 
        else 
                printf("f is NOT 1.0
"); 

        return 0; 
} 

它总是打印 f is NOT 1.0.我知道这与 C 中的浮点精度有关.但我不确定它到底在哪里搞砸了.有人可以解释一下为什么它不打印另一行吗?

It always prints f is NOT 1.0. I understand this is related to floating point precision in C. But I am not sure exactly where it is getting messed up. Can someone please explain me why it is not printing the other line?

推荐答案

二进制浮点数不能准确表示值 0.1，因为它的二进制展开没有有限的位数(和十进制展开完全一样)1/7 没有).

Binary floating point cannot represent the value 0.1 exactly, because its binary expansion does not have a finite number of digits (in exactly the same way that the decimal expansion of 1/7 does not).

0.1的二进制展开是

0.000110011001100110011001100...

当截断为 IEEE-754 单精度时，这大约是十进制的 0.100000001490116119.这意味着每次将接近 0.1"的值添加到变量时，都会累积一个小错误 - 因此最终值略高于 1.0.

When truncated to IEEE-754 single precision, this is approximately 0.100000001490116119 in decimal. This means that each time you add the "nearly 0.1" value to your variable, you accumulate a small error - so the final value is slightly higher than 1.0.

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