为什么不能打印出链表的值，请指教

Question

#include #include struct node{ int data; struct node*next; }; struct node*head; struct node*insert(int x){ struct node*a=(struct node*)malloc(sizeof(struct node)); a->data=x; a->next=NULL; return a; }; void insertathead(int x){ struct node*b=insert(x); if(head==NULL) { head=b; return; } b->next=head; head=b; }; void print(){ struct node*c=head; while(c->next!=NULL); { printf("%d\t",c->data); c=c->next; } return; }; int main (void) { head=NULL; insertathead(1); insertathead(2); insertathead(3); print(); return 0; }

Answer 1

循环条件改为while (c != NULL)，可以打呀。你看下面的代码和运行结果。

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

struct node {
    int data;
    struct node * next;
};

struct node * head;
struct node * insert(int x) {
    //printf("Creating node for data %d\n", x);
    struct node * a = (struct node * ) malloc(sizeof(struct node));
    a -> data = x;
    a -> next = NULL;
    return a;
};

void insertathead(int x) {
    struct node * b = insert(x);
    if (head == NULL) {
        //printf("Head is NULL. %d is head.\n", x);
        head = b;
        return;
    }
    //printf("Head is not NULL. %d is inserted to head.\n", x);
    b -> next = head;
    head = b;
};

void print() {
    struct node * c = head;
    while (c != NULL) {
        printf("%d\t", c -> data);
        c = c -> next;
    }
    return;
};

int main(void) {
    head = NULL;
    insertathead(1);
    insertathead(2);
    insertathead(3);
    print();
    return 0;
}



// Output
3       2       1