#include <stdio.h>
#include <string.h>
typedef struct test_s test_t;
struct test_s {
int a;
int b;
char arr[0];
};
int main()
{
test_t *t;
char buf[32] = {0};
int i;
t = (test_t *)buf;
t->a = 1;
t->b = 2;
memcpy(t->arr, "123", 3);
printf("%d, %d, %s\n", sizeof(test_t), t->a, t->arr);
for (i = 0; i < sizeof(buf); i++) {
printf("%x ", buf[i]);
}
printf("\n");
return 0;
}
运行结果
8, 1, 123
1 0 0 0 2 0 0 0 31 32 33 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
如果做下调整
struct test_s {
int a;
char arr[0];
int b;
};
则会导致覆盖
1 0 0 0 31 32 33 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
说明 char arr[0],没有静态分配内存,只是定义一个符号给编译器用,符号表示的地址为 char arr[0] 前声明的分配了内存的空间的偏移。
这样的好处:避免指针的复杂操作,比如上面代码用指针只能:
typedef struct test_s test_t;
struct test_s {
int a;
int b;
char *arr;
};
int main()
{
test_t *t;
char buf[32] = {0};
int i;
t = (test_t *)buf;
t->a = 1;
t->b = 2;
t->arr = (char *)t + sizeof(*t);
memcpy(t->arr, "123", 3);
printf("%d, %d, %s\n", sizeof(test_t), t->a, t->arr);
for (i = 0; i < sizeof(buf); i++) {
printf("%x ", buf[i]);
}
printf("\n");
return 0;
}
运行结果
12, 1, 123
1 0 0 0 2 0 0 0 ffffffa8 69 ffffffcb ffffffbf 31 32 33 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
标签:arr,int,char,数组,printf,test,动态,buf
From: https://www.cnblogs.com/yangxinrui/p/16899172.html