为什么会出现这种情况，如果失败的消极和积极的整数比较 [英] Why does this if condition fail for comparison of negative and positive integers

问题描述

#include <stdio.h>

int arr[] = {1,2,3,4,5,6,7,8};
#define SIZE (sizeof(arr)/sizeof(int))

int main()
{
        printf("SIZE = %d\n", SIZE);
        if ((-1) < SIZE)
                printf("less");
        else
                printf("more");
}

与 GCC 编译后的输出是更多。为什么如果一条件，则即使 -1 LT; 8 ？

The output after compiling with gcc is "more". Why the if condition fails even when -1 < 8?

推荐答案

问题是在你的比较：

    if ((-1) < SIZE)

的sizeof 通常返回无符号长，所以尺寸将无符号长，而 1 只是一个 INT 。在C和相关语言晋升的规则意味着-1将在比较之前被转换成为size_t ，所以 1 将成为一个非常大的正值（一个无符号长）。

sizeof typically returns an unsigned long, so SIZE will be unsigned long, whereas -1 is just an int. The rules for promotion in C and related languages mean that -1 will be converted to size_t before the comparison, so -1 will become a very large positive value (the maximum value of an unsigned long).

解决这个问题的方法之一是改变比较：

One way to fix this is to change the comparison to:

    if (-1 < (long long)SIZE)

但它实际上是一个毫无意义的比较，因为一个无符号值总是> = 0定义，编译器很可能就此发出警告。

although it's actually a pointless comparison, since an unsigned value will always be >= 0 by definition, and the compiler may well warn you about this.

随后被@Nobilis注意，你应启用编译器警告和注意他们：如果您有例如编译的gcc -Wall ... 编译器会提醒你你的bug。

As subsequently noted by @Nobilis, you should always enable compiler warnings and take notice of them: if you had compiled with e.g. gcc -Wall ... the compiler would have warned you of your bug.

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