C 设置一个指向另一个指针的指针并释放

Question

如果我调用free(ptr)，p1也会被释放吗？

int* some_function(){
    int *p1 = malloc(sizeof(int)*10);
    int *p2 = p1;
    return p2;
}
int main(){
    int *ptr = some_function();
    free(ptr);
    return 0;
}

Answer 1

当你这样做时

int *p1 = malloc(sizeof(int)*10);

你有类似的东西

+----+     +---------------------------------+
| p1 | --> | Memory enough for 10 int values |
+----+     +---------------------------------+

Then when you do

int *p2 = p1;

你有类似的东西

+----+
| p1 | --\
+----+    \     +---------------------------------+
           >--> | Memory enough for 10 int values |
+----+    /     +---------------------------------+
| p2 | --/
+----+

That is, you have two pointers pointing to the very same memory.

When you return p2 you simply copy the pointer, much like what happens when you initialize p2, and inside the main function you have

int *ptr = some_function();

这会导致

+-----+     +---------------------------------+
| ptr | --> | Memory enough for 10 int values |
+-----+     +---------------------------------+

这是不同的变量，但它们的内容都是相同的，并且都是初始malloc调用返回的指针。