小编Kri*_*dar的帖子

#include<stdio.h>
int main() {
int buff[] = {1,2,3,4,5,6,9,10};
char c = (buff+1)[5];
printf("%d\n",c);//output is 9
return 0;
}

有人可以清楚地解释这是如何发生的以及为什么

#include<stdio.h>
#include<stdlib.h>
int main( int argc ,char** argv) {
    int bugs = 100;
    char nul_byte='\0';
    char care_percentage = bugs * nul_byte;
    printf("Which means you should care %s%%.\n",care_percentage);// 1->prints (null)
    printf("Which means you should care %d%%.\n",care_percentage);// 2->prints 0
    printf("Which means you should care %c%%.\n",care_percentage);// 3->prints 
    return 0;
}

问题是在最后三个语句1,2和3中发生了什么.在机器级内部发生了什么.%s如何(null)和%d将其视为0,%c在打印时将其视为空.

有人可以在机器深度解释这些吗？