递归打印链接列表时出现分段错误

Question

我在C中编写了一个链表(使用gcc编译器),并试图以递归方式打印它.它告诉"分段错误",并且还仅打印第一个值.任何人都可以提出纠正它的选项..？这是我的代码.

#define MAX 10
#include <stdio.h>
#include <stdlib.h>
struct node {
    int value;
    struct node *next;
};

void printRecursively(struct node *start) {
    if (start != NULL) {
        printf("%d\n", start->value);
        start = start->next;
        printRecursively(start);
    }
}

void main() {
    struct node *nodes[MAX];
    for (int i = 0; i < MAX; i++) {
        nodes[i] = malloc(sizeof(struct node));
        nodes[i]->value = i + 1;
        nodes[i]->next = nodes[i + 1];
    }
    printRecursively(nodes[0]);
}

Answer 1

您的代码将next每个新分配的代码的指针初始化为未初始化的值.向后运行循环并确保初始化ast节点的next指针NULL.

int main(void) {
    struct node *nodes[MAX];
    for (int i = MAX; i-- > 0;) {
        nodes[i] = malloc(sizeof(struct node));
        nodes[i]->value = i + 1;
        nodes[i]->next = (i == MAX - 1) ? NULL : nodes[i + 1];
    }
    printRecursively(nodes[0]);

    /* for good style, free the allocated memory */
    for (int i = 0; i < MAX; i++) {
        free(nodes[i]);
    }
    return 0;
}

如您所述,有一个简单的解决方案,即增加索引值并检查内存分配失败:

int main(void) {
    struct node *nodes[MAX];
    for (int i = 0; i < MAX; i++) {
        nodes[i] = malloc(sizeof(struct node));
        if (nodes[i] == NULL) {
            fprintf(stderr, "memory allocation failure\n");
            exit(1);
        }
        nodes[i]->value = i + 1;
        nodes[i]->next = NULL;
        if (i > 0) {
            nodes[i - 1]->next = nodes[i];
        }
    }
    printRecursively(nodes[0]);

    /* for good style, free the allocated memory */
    for (int i = 0; i < MAX; i++) {
        free(nodes[i]);
    }
    return 0;
}