相关疑难解决方法(0)

#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;
}

上面编译的程序的输出gcc是

0
1
1

使用-Wall或-Waddress选项,gcc发出警告:

warning: the address of ‘i’ will always evaluate as ‘true’ [-Waddress]

如何c在上述计划中进行评估？