#include <stdio.h>

int main(int argc, const char * argv[]) {
   
    int a[ ] ={1,2,3,4,5};
    int *p = (int*)(&a + 1);
    printf("%d \n %d \n",*(a + 1),*(p - 1));
}
1.请问打印的结果是什么?

1. int a[ ] ={1,2,3,4,5};
   
2.     int *p = (int*)(&a + 1);
   
3.     printf("%d \n %d \n",*(a + 1),*(p - 1));
   
4.     printf("-----------------------\n");
   
5.     printf("%p\n",a);
   
6.     printf("%p\n",&a);
   
7.     printf("%p\n",a+1);
   
8.     printf("%p\n",a+2);
   
9.     printf("%p\n",a+3);
   
10.     printf("%p\n",a+4);
    
11.     printf("%p\n",&a+1);
    
12.     printf("%p\n",p-1);
    
13.     printf("%p\n",p);
    
14.     printf("%lu\n",sizeof(int*));
    
15.     printf("%p\n",(int*)1);
    

复制代码

结果如下：2
5
-----------------------
0x7fff5fbff770
0x7fff5fbff770
0x7fff5fbff774
0x7fff5fbff778
0x7fff5fbff77c
0x7fff5fbff780
0x7fff5fbff784
0x7fff5fbff780
0x7fff5fbff784
8
0x1
Program ended with exit code: 0
&a+1越界了，*（a+1）是数组的第二个元素，跟指针p没有关系

的确 是 2和5

这题好有迷惑性~~~

应该是 2 ---- 5

同楼上！

2     5
&a+1越界,P-1取回第五个元素地址*一下取值得5

&a+1 确实涨知识，搞清楚了就知道为什么是5。谢谢您~~

2   5吗？？？