i = &i

咕噜咕噜斯基

于 2012-09-27 20:28:00 发布

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分类专栏： C/C++ 文章标签：编译器

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// 偶然的在网上看到这么个问题, 问下列程序有没有一错误. 
#include <stdio.h>

int main(void)
{
    int i;
    printf("int i;\ni == %d\naddr == %d\n\n", i, &i);

    i = (int)&i;
    printf("i = &i;\ni == %d\naddr == %d\n\n", i, &i);

    scanf("%d", i);
    printf("i == %d\naddr == %d", i, &i);

    return 0;
}

本来以为编译器会报出错误. 而实际编译时, Code::Blocks 只是显示下列警告:

// 警告：格式 ‘%d’ expects argument of type ‘int *’, but argument 2 has type ‘int’ [-Wformat]

输出结果:

int i;
i == 2147340288
addr == 2293532


i = &i;
i == 2293532
addr == 2293532


123456
i == 123456
addr == 2293532
Process returned 0 (0x0)   execution time : 3.231 s
Press any key to continue.

记录一下表示学习!!!! 长长见识!!!!!!

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