Question

所以我在一个单独的函数中创建一个链表，当我打印出函数中的链表时，似乎一切都很好。然而;当我去主要并尝试使用printf访问链接列表时，我得到了一个分段错误，我很困惑。

void createLL(struct node* head, struct node* curr, char ch, int number){
//lowest digit is the head
    while (((scanf(" %c",&ch)) >= 0)){
        curr = (struct node*)malloc(sizeof(struct node*)); //allocate space
        number = ch - '0' ; //convert char to number
        curr->data = number;
        curr->next = head;
        head = curr;
    }
    curr = head;
    //troubleshoot
    while(curr){
        printf("%d\n",curr->data);
        curr = curr->next;
    }
    curr = head;
    printf("%d\n",curr->data);
}

int main(){
    //initials
    int i, number; 
    char ch;
    //node pointers
    struct node* headOne = NULL;
    struct node* currOne = NULL;
    struct node* headTwo = NULL;
    struct node* currTwo = NULL;
    //create linked list
    createLL(headOne,currOne, ch, number);
    printf("%d\n",currOne->data);
    createLL(headTwo,currTwo, ch, number);
    printf("%d\n",currTwo->data);

Answer 1

在C函数中，按值传递所有参数。因此，如果要更改函数中的变量，则需要传递该变量的地址并取消引用该函数中的参数。

此外，您没有为节点分配适当的空间。您想要sizeof(struct node)，而不是sizeof(struct node *)。

void createLL(struct node **head, struct node **curr, char ch, int number){
//lowest digit is the head
    while (((scanf(" %c",&ch)) >= 0)){
        // don't cast the return value of malloc
        *curr = malloc(sizeof(struct node)); //allocate space
        number = ch - '0' ; //convert char to number
        (*curr)->data = number;
        (*curr)->next = *head;
        *head = *curr;
    }
    *curr = *head;
    //troubleshoot
    while(*curr){
        printf("%d\n",(*curr)->data);
        *curr = (*curr)->next;
    }
    *curr = *head;
    printf("%d\n",(*curr)->data);
}


int main(){
    //initials
    int i, number; 
    char ch;
    //node pointers
    struct node* headOne = NULL;
    struct node* currOne = NULL;
    struct node* headTwo = NULL;
    struct node* currTwo = NULL;
    //create linked list
    createLL(&headOne,&currOne, ch, number);
    printf("%d\n",currOne->data);
    createLL(&headTwo,&currTwo, ch, number);
    printf("%d\n",currTwo->data);
}

创建链接列表，而不是传回主要

1 个答案: