关于数组两个元素地址相减的问题

#include<stdio.h>

int a[5]={1,2,5,9,8};

main(){
int m,n,l,q;
m = a-&a[3];
n = (char*)a-(char*)&a[3];
l = (int)a - (int)&a[3];
q = (int*)a - (int*)&a[3];

//&a[3] = a + 3;

printf("Test the value of a is %d\n",a);
printf("Test the value of &a[0] is %d \n",&a[0]);
printf("Test the value of &a[1] is %d \n",&a[1]);
printf("Test the value of &a[2] is %d \n",&a[2]);
printf("Test the value of &a[3] is %d \n",&a[3]);

printf("Test the value of m is %d \n",m);
printf("Test the value of n is %d \n",n);
printf("Test the value of l is %d \n",l);
printf("Test the value of q is %d \n",q);
}

　　

运行结果：



查看反汇编的代码，发现：
int nTmp = &a[4] - &a[0];
00416B87 lea eax,[ebp-28h]
00416B8A lea ecx,[arrayTmp]
00416B8D sub eax,ecx
00416B8F sar eax,2
00416B92 mov dword ptr [nTmp],eax
原来，执行完数组地址相减运算后，还会执行算数右移指令，右移位数视参数类型而定，如int型右移2位，short型右移1位。都知道右移1位相当于除以2操作，右移2位等同于除以4。

由此可见，两个数组元素地址相减，实际是获取两个元素数组元素的距离，而不是地址的距离。如果要计算地址距离，就直接强制类型转换：int nTmp = (char*)&a[4] - (char*)&a[0];