代码:
#include<stdio.h>typedef struct test{int a;int b;int c;} TEST;TEST arr[]={{1,2,3},{4,5,6},{7,8,9}};void main(void){TEST *tmp;tmp = arr;printf("tmp: %d, %d, %d,%d, %d, %d,%d, %d, %d\n", tmp[0].a, tmp[0].b, tmp[0].c, tmp[1].a, tmp[1].b, tmp[1].c, tmp[2].a, tmp[2].b, tmp[2].c);printf("arr: %d, %d, %d,%d, %d, %d,%d, %d, %d\n", arr[0].a, arr[0].b, arr[0].c, arr[1].a, arr[1].b, arr[1].c, arr[2].a, arr[2].b, arr[2].c);}
执行结果:
[baoliw@AONTFN07 ~]$ ./a.out
tmp: 1, 2, 3,4, 5, 6,7, 8, 9
arr: 1, 2, 3,4, 5, 6,7, 8, 9
结果分析:
1)数组名可以直接赋值给相同类型的指针变量
2)指针变量可以采用类似数组的用法,如:tmp[0].a
注:不能将数组名赋值给数组变量,如:
TEST jj[3];
jj = arr;
编译将产生错误: error: incompatible types when assigning to type ‘struct TEST[3]’ from type ‘struct TEST *’