为什么sizeof类型与整数比较返回false [英] why sizeof type compared with integer returns false

问题描述

我在下面的代码中发现了一种奇怪的行为.

one strange behaviour I noticed in below code.

#include<stdio.h>
#include <stdbool.h>

int main()
{
int x = sizeof(int) > -1;

bool z = sizeof(int);

printf("x is %d \t z is %d \n",x,z);
if(sizeof(int)>-1)
{
printf("true\n");
}
else
printf("false\n");
}

当 sizeof(int)> -1 为true且预期输出应为1时，为什么int x为零.

Why int x is zero when sizeof(int) > -1 is true and the expected output should be 1.

推荐答案

sizeof 运算符产生的不是 int ，而是产生的 size_t 无符号整数类型.当您将有符号整数(如-1)与无符号整数进行比较时，最终将比较错误的值.

The sizeof operator yields not an int but size_t which is an unsigned integer type. When you compare a signed integer like -1 to an unsigned integer you will end up comparing the wrong values.

进行以下更改，代码将按预期工作.

Do the following changes and the code will work as expected.

#include<stdio.h>

#include <stdbool.h>

int main()
{
    int x = (int)sizeof(int) > -1;

    bool z = sizeof(int);

    printf("x is %d \t z is %d \n",x,z);
    if((int)sizeof(int) > -1)
    {
        printf("true\n");
    }
    else
        printf("false\n");
}

输出:

x is 1   z is 1
true

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