c - 如何从另一个函数释放一个函数动态分配的内存？

Question

在我附加的 C 程序中，我定义了一个名为 push() 的单独函数，用于将节点添加到链表的前面。 push() 为堆上的 node 分配内存，但我无法释放此处的内存，因为这样 push() 完成的工作将不会反射(reflect)在调用者中 (main())。那么如何从 main() 内部释放相关的堆分配内存呢？

感谢任何形式的帮助。提前致谢。

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

struct node
{
    int data;
    struct node *next;
};

/* Prototypes */
void push(struct node **headRef, int data);

int main(void)
{
    struct node *head, *tail, *current;
    int i;

    head = NULL;

    // Deal with the head node here, and set the tail pointer
    push(&head, 1);
    tail = head;        // tail and head now point to the same thing

    // Do all the other nodes using TAIL
    for (i = 2; i < 6; i++)
    {
        push(&(tail->next), i);     // add node at tail->next
        tail = tail->next;          // advance tail to point to last node
    }

    current = head;
    while (current)
    {
        printf("%d ", current->data);
        current = current->next;
    }
    printf("\n");

    return 0;
}

/*
 Takes a list and a data value.
 Creates a new link with the given data and pushes
 it onto the front of the list.
 The list is not passed in by its head pointer.
 Instead the list is passed in as a "reference" pointer
 to the head pointer -- this allows us
 to modify the caller's memory.
*/
void push(struct node **headRef, int data)
{
    struct node *newNode = malloc(sizeof(struct node));
    newNode->data = data;
    newNode->next = *headRef;   // The '*' to dereference back to the real head
    *headRef = newNode;         // ditto
}

Answer 1

您可以像这样释放main中分配的空间 -

struct node * tmp;
while(head){
    tmp = head;
    head = head->next; //this is to avoid loosing reference to next memory location
    free(tmp); 
}

由于您在 push 中传递了变量的地址，因此这是可能的。