C-释放指针后，该指针不为null

Check it out here:C - How can I free dynamically allocated memory?List* head1 = NULL;insertFront(&head1, 1);insertFront(&head1, 2);print(head1);while (head1){ List *temp = head1; head1 = head1->next; free(temp);}if(head1 == NULL) printf("Null\n");else printf("Not null\n");在这种情况下的输出:Null在这种情况下，在释放head1(也有节点)之后，head1变为null，不是吗?In this case after freeing head1 (nodes also) the head1 becomes null, does'nt it?最后，我是否缺少一些概念?And finally, am I missing some concepts? head1为空，但p不是.head1 is null, however p isn't.我的问题是: 为什么head1和p之间的值不同?推荐答案循环条件while (head1) ^^^^^^{ List *temp = head1; head1 = head1->next; free(temp);}当下一个节点的数据成员next等于NULL时，等于falsebecomes equal to false when the data member next of the next node is equal to NULLhead1 = head1->next;^^^^^ ^^^^^这与功能free没有共同之处.该函数按值接受参数.那就是它处理原始[指针的副本.因此原始指针本身不会被更改.This has nothing common with the function free. The function accepts the argument by value. That is it deals with a copy of the original [pointer. So the original pointer itself will not be changed. 这篇关于C-释放指针后，该指针不为null的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！ 上岸，阿里云！