-1

我正在玩链接列表以适应它,但我无法让这个小程序工作。我不知道这里有什么问题,求助。

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
//Struct
struct Node {
    int value;
    struct Node *next;
};
typedef struct Node NODE;

//Function Declaration
NODE* addNode (NODE* pList, NODE* pPre, int data);
void printList (NODE* pList);

int main (void)
{
    //Local Declaration
    NODE *pPre;
    NODE *pList;

    //Statement
    pList = addNode (pList, pPre, 10);
    pList = addNode (pList, pPre, 20);
    pList = addNode (pList, pPre, 30);

    printList (pList);
    return 0;
}

NODE* addNode (NODE* pList, NODE* pPre, int data)
{
    //Local Declaration
    NODE* pNew;

    //Statement
    if (!(pNew = (NODE*)malloc(sizeof(NODE))))
    {
        printf("\aMemory overflow in insert\n");
        exit(1);
    }

    pNew->value = data;
    if (pPre == NULL)
    {
        //Inserting before first node or to empty list.
        pNew->next = pList;
        pList = pNew;
    }
    else
    {
        pNew->next = pPre->next;
        pPre->next = pNew;
    }
    return pList;
}
void printList (NODE* pList)
{
    //Local Declaration
    NODE* pNew;

    //Statement

    pNew = pList;
    while(pNew)
    {
        printf("%d", pNew->value);
        pNew = pNew->next;

    }
    return;
}

pPre 是前驱节点,pList 是指向列表的指针。

4

2 回答 2

2

您尚未将 NULL 分配给指针pPrePList,请尝试以下代码,它现在运行正常,

NODE *pPre=NULL;
NODE *pList=NULL;
于 2013-03-23T05:12:20.187 回答
0

根据您的代码逻辑,您希望 addNode() 修改 pPre。所以你需要将 pPre 定义为 Node * 的指针。

Node * addNode(Node *pList, Node **pPre, int data)

addNode() 应该返回 pList,你可能会错过它。

于 2013-03-23T04:55:52.923 回答