我的电脑认为有符号整数比-1小?

#include <stdio.h>
int main(void)
{
    printf("%d", sizeof(signed int) > -1);
    return 0;
}

the result is 0 (FALSE). how can it be? Im using 64bit ubuntu linux so the result should be (4 > -1) => 1 => True.

结果是0 (FALSE)。为什么会这样呢?我使用64位ubuntu linux，所以结果应该是(4 > -1)=> 1 =>为真。

3 个解决方案

#1

sizeof(signed int) has type size_t, which is an unsigned type. When you make a comparison between a signed and an unsigned value [and the unsigned value's type is at least as large as the signed value's type], the signed value is converted to unsigned before the comparison. This conversion causes -1 to become the largest possible value of the unsigned type. In other words, it is as if you wrote

sizeof(带符号int)具有类型size_t，这是一个无符号类型。当您在一个有符号的值和一个无符号的值之间进行比较时[且无符号的值的类型至少与有符号的值的类型一样大]，在比较之前将有符号的值转换为无符号的值。这种转换使-1成为无符号类型的最大可能值。换句话说，就好像你写了一样

#include <limits.h>
/* ... */
printf("%d", sizeof(signed int) > SIZE_MAX);

You can make gcc warn when you make this mistake, but it's not on by default or even in -Wall: you need -Wextra or more specifically -Wsign-compare. (This warning can produce a great many false positives, but I think it's useful to turn on for new code.)

当您犯这个错误时，您可以发出gcc警告，但它不是默认的，甚至不是在-Wall中:您需要-Wextra或更具体地-Wsign-compare。(这个警告可能会产生大量的误报，但我认为打开新代码是有用的。)

3 个解决方案

#1

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