0

这有效:

struct list
{
    int value;
    struct list *next;
};

void addElement(struct list **l, int value)
{
    if( *l == 0 )
    {
        *l = (struct list *) malloc(sizeof(struct list));
        (*l)->value = value;
        (*l)->next = 0;
    }
    else
    {
        struct list *new_element = (struct list *) malloc(sizeof(struct list));
        new_element->value = value;
        new_element->next = *l;
        *l = new_element;
    }
}

void printList(struct list *l)
{
    struct list *temp;
    for(temp = l;temp; temp = temp->next)
    printf("%d ->", temp->value);
    printf("%d",0);
}

int main(int argc, char **argv)
{
    printf("Einfache verkettete Liste: ");
    struct list *mylist;    
    int i;

    addElement(&mylist,10);
    addElement(&mylist,10);
    addElement(&mylist,10);
    addElement(&mylist,10);
    addElement(&mylist,10);
    addElement(&mylist,10);
    addElement(&mylist,10);
    addElement(&mylist,10);
    addElement(&mylist,10);
    addElement(&mylist,10);

    printList(mylist);

    return 0;
}

输出:

Einfache verkettete Liste: 10 ->10 ->10 ->10 ->10 ->10 ->10 ->10 ->10 ->10 ->0

但如果我改变:

addElement(&mylist,10);
addElement(&mylist,10);
addElement(&mylist,10);
addElement(&mylist,10);
addElement(&mylist,10);
addElement(&mylist,10);
addElement(&mylist,10);
addElement(&mylist,10);
addElement(&mylist,10);
addElement(&mylist,10);

至:

for(i=0;i<10;i++)
    addElement(&mylist,10);

它在运行时出现内存错误?它非常混乱,我不知道发生了什么。如何调试?

4

1 回答 1

8

您从未初始化mylist为 null,因此在您第一次调用addElement时,是否*l == 0为真是运气问题。看似无关的代码重新排列会改变你的运气。

于 2013-09-17T17:57:04.357 回答