1.题目:
void main(){ int a ,b,k=4,m=6,*p1=&k,*p2=&m;
a=(p1==p2);
b=(*p1)/(*p2)+7; printf("a=%d\n,b=%d\n",a,b);}
2.代码+分析:
#include <stdio.h>
void main() {
int a, b, k = 4, m = 6, *p1 = &k, *p2 = &m;
a = (p1 == p2);
b = (*p1) / (*p2) + 7;
printf("a=%d,b=%d\n", a, b);
//分析:指针变量p1指向变量k,指针变量p2指向变量m
//首先a = (p1 == p2),p1和p2不相等,所以a=0
//然后b = (*p1) / (*p2) + 7,b=4/6+7=7
//输出a=0,b=7
}