#include <stdio.h>

volatile int i;

int main()
{
    int c;

    for (i = 0; i < 3; i++) 
    {
         c = i &&& i;
         printf("%d\n", c);
    }

    return 0;
}

使用编译的上述程序的输出gcc是

0
1
1

使用-Wallor-Waddress选项，gcc发出警告：

warning: the address of 鈥榠鈥� will always evaluate as 鈥榯rue鈥� [-Waddress]

c在上述计划中如何评估？