内存分配释放
我目前正在编写一个链表，并尝试在删除节点时释放内存分配。然而，经过几个小时的尝试，我似乎无法获得干净的 valgrind 输出。
void * pop(struct List *list)
{
if(list->head == 0){
return 0;
}
struct Node * tempNode = list->head->next;
free(list->head);
list->head = tempNode;
...
}
我通过以下方式分配空间：
addNode(struct List *list, void *element){
struct Node *node;
node = (struct Node *)malloc(sizeof(node));
....
}
基本上在 pop 函数中，我想取出列表的头部并使头部的下一个节点成为新的头部。我想释放分配给 head 的内存。
感谢您的帮助
I'm currently writing a linked list and trying to free up memory allocations when I delete a node. However, after hours of trying to do this, I can't seem to get a clean valgrind output.
void * pop(struct List *list)
{
if(list->head == 0){
return 0;
}
struct Node * tempNode = list->head->next;
free(list->head);
list->head = tempNode;
...
}
I'm allocating the space by saying:
addNode(struct List *list, void *element){
struct Node *node;
node = (struct Node *)malloc(sizeof(node));
....
}
Basically in the pop function I want to take out the head of the list and make the head's next node the new head. I want to deallocate the memory that was given to head.
Thanks for any help
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评论（3）
哇哦，你的
malloc不正确。你有：你需要的是：
在你的原始代码中，你只为指针分配了足够的空间。但是您正在尝试分配一个 Node 对象。
Woah, your
mallocisn't correct. You have:What you need is:
In your original code, you are only allocating enough for a pointer. But you are trying allocate a
Nodeobject.为
node指向的事物分配空间。不要强制转换
malloc的返回值。这样做可以掩盖#include的失败。Allocate space for the thing pointed to by
node.Don't cast the return value of
malloc. Doing so can mask the failure to#include <stdlib.h>.看来是对的。还是其他地方有漏水的情况？或者您是否将元素本身作为传入的参数释放？
It seems correct. Or are there some leaks in the other places? Or do you free the element itself as the parameter passed in ?