为什么如果条件无法比较负整数和正整数

时间:2023-03-21
本文介绍了为什么如果条件无法比较负整数和正整数的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

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
", SIZE);
        if ((-1) < SIZE)
                printf("less");
        else
                printf("more");
}

gcc编译后的输出是"more".为什么即使 -1 -1 条件也会失败?8?

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

推荐答案

问题在于你的比较:

    if ((-1) < SIZE)

sizeof 通常返回一个 unsigned long,所以 SIZE 将是 unsigned long,而 -1 只是一个 int.C及相关语言的提升规则是-1会在比较前转换为size_t,所以-1会变成一个非常大的正值(最大值unsigned long).

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).

解决此问题的一种方法是将比较更改为:

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

    if (-1 < (long long)SIZE)

尽管这实际上是一个毫无意义的比较,因为根据定义,无符号值总是 >= 0,并且编译器很可能会就此警告您.

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.

正如@Nobilis 随后指出的那样,您应该始终启用编译器警告并注意它们:如果您使用例如编译gcc -Wall ... 编译器会警告你你的错误.

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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