这段 C 代码底层发生了什么？

[lhx2008]

name 是否是在离开作用域时由编译器重新分配了内存？是在哪个区域分配了空间？这样的行为是未定义的吗？

#include <stdlib.h>
#include <stdio.h>

typedef struct {
  const char*  name;
} island;

int main() {
  island *in_heap =  (island*) malloc(sizeof(island));
  {
    char name[80];
    fgets(name, 80, stdin);
    in_heap->name = name;
    fgets(name, 80, stdin);
    printf("address of name in scope = %p \n", &name);
  }
  printf("\nout of scope \n");
  printf("name is not deleted, name = %s", in_heap->name);
  printf("address of in_heap->name = %p \n  ", &(in_heap->name));
  free(in_heap);
}

输入

test
test2

输出

address of name in scope = 0x7ffffce79b10

out of scope
name is not deleted, name = test2
address of in_heap->name = 0x7ffff5954260

[wwqgtxx]

这种算非法访问堆内存吧，具体行为应该是未定义的

[wwqgtxx]

@wwqgtxx 堆内存->栈内存

[wwqgtxx]

而且你%p 的时候再取一次地址干嘛

[lhx2008]

@wwqgtxx 是，这里写的时候没想明白，应该还在栈上

[haiyang416]

```
printf("address of in_heap->name = %p \n ", in_heap->name);
```