C++ 为什么说双重免费或腐败(出)已中止

发布于07月15日

我写了这段代码，它说有错误double free or corruption(out)，我真的不明白我在哪里搞砸了.

int main(void) {
    node *list = NULL;
    node *n = malloc(4 * sizeof(node));
    if (n == NULL) {
        return 1;
    }

    n[0].number = 1;
    n[0].next = NULL;
    list = &n[0];

    n[1].number = 2;
    n[1].next = NULL;
    list->next = &n[1];

    n[2].number = 3;
    n[2].next = NULL;
    list->next->next = &n[2];

    n[3].number = 4;
    n[3].next = NULL;
    list->next->next->next = &n[3];

    for (node *tmp = list; tmp != NULL; tmp = tmp->next) {
        printf("%i\n", tmp->number);
    }

    while (list != NULL) {
        node *tmp = list->next;
        free(list);
        list = tmp;
    }
}

推荐答案

如果将 node 分配为由4个 node 组成的单个块，则无法像在最终循环中那样一次释放一个 node .您应该:

分别分配每个 node
或者仅释放由n指向的 node array.

以下是使用第一种方法的修改版本:

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

typedef struct node {
    int number;
    struct node *next;
} node;

node *new_node(int number) {
    node *n = malloc(sizeof(*n));
    if (n == NULL) {
        perror("new_node");
        exit(1);
    }
    n->number = number;
    n->next = NULL;
    return n;
}

int main(void) {
    node *list;

    list = new_node(1);
    list->next = new_node(2);
    list->next->next = new_node(3);
    list->next->next->next = new_node(4);

    for (node *tmp = list; tmp != NULL; tmp = tmp->next) {
        printf("%i\n", tmp->number);
    }

    while (list != NULL) {
        node *tmp = list->next;
        free(list);
        list = tmp;
    }
    return 0;
}