Question

我正在尝试创建程序来构建链接列表，但是在创建第二个错误时它会给我一个分段错误。

[root @ vm c_prog] #vi link1.c

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


struct node {

        int x;
        struct node *next;
        };

int main () {

        int d;
        struct node *root;
        struct node *current;



        root = malloc(sizeof(struct node));

        current = root;

        printf ("Location of root is %p \n", root);

d = 1;
while (d>0){

        printf ("Enter the value of X: ");
        scanf ("%d", &current->x);

        printf ("value of x is stored\n");
        printf ("Location of current is %p \n", current);
        printf ("Value of X in 1st node: %d\n", current->x);

        current = current->next;


        printf ("Enter zero to terminate the loop: ");

        scanf ("%d",&d);

        }

}

[root@vm c_prog]# ./a.out 
Location of root is 0xc34b010 
Enter the value of X: 3
value of x is stored
Location of current is 0xc34b010 
Value of X in 1st node: 3
Enter zero to terminate the loop: 5
Enter the value of X: 5
Segmentation fault
[root@vm c_prog]#

Answer 1

您永远不会初始化next所以行

current = current->next;

将当前值更改为指向未初始化的内存。您还需要为循环的每次迭代分配一个新的node。

以下代码应该接近工作。（它可以简化;我试图让它尽可能接近你的代码。）

int main () {
    int d;
    struct node *root = NULL;
    struct node *current;
    d = 1;
    while (d>0){
        printf ("Enter the value of X: ");
        scanf ("%d", &d);
        if (root == NULL) {
            root = calloc(1, sizeof(struct node));
            current = root;
            printf ("Location of root is %p \n", root);
        }
        else {
            current->next = calloc(1, sizeof(struct node));
            current = current->next;
        }
        current->x = d;

        printf ("value of x is stored\n");
        printf ("Location of current is %p \n", current);
        printf ("Value of X in last node: %d\n", current->x);

        printf ("Enter zero to terminate the loop: ");

        scanf ("%d",&d);

        }
}

Answer 2

除了simonc所说的（这是正确的）之外，您应该将next设置为循环中的某些内容。

链接列表程序在构建下一个节点时崩溃

2 个答案: