Question

As always being this problem comes from a book s exercises. Neither I am not studying data structures, nor the book is about that. But, there is a chapter that is "Dynamic Data Structures". I ve already finished this chapter.

但是,我对插入有异议。在我看来,我的功能是正常的,除了它产生重复的节点之外。

I made a precaution for that but it doesn t work. Anyway, please forgive me because of my silly mistakes. OK, Here my structure types for name lists.

typedef struct name_node_s {
char name[11];
struct name_node_s *restp;
}name_node_t;

typedef struct {
    name_node_t *headp;
    int size;
}name_list_t;

位置_ 第一个函数 :

name_node_t *
place_first(name_list_t *old_listp, char name[11])
{
    name_list_t *new_listp, *cur_listp;
    name_node_t *new_nodep, *temp_nodep;
    temp_nodep = (name_node_t *)malloc(sizeof (name_node_t));
    new_listp = (name_list_t *)malloc(sizeof (name_list_t));
    cur_listp = (name_list_t *)malloc(sizeof (name_list_t));
    new_nodep = (name_node_t *)malloc(sizeof (name_node_t));
    cur_listp->headp = old_listp->headp;
    temp_nodep = old_listp->headp;
    new_listp = old_listp;
    if ( old_listp->headp->name != name ){ // My first precaution for duplication
        while(cur_listp->headp->restp != NULL && cur_listp->headp->name != name) // My second precaution for duplication
        {
            if (old_listp->headp == NULL){
                strcpy(new_listp->headp->name, name);
                new_listp->headp->restp = NULL;
            }
            else if (old_listp->headp->name != name) { // Third one.
                strcpy(new_nodep->name, name);
                new_nodep->restp = NULL;
                new_listp->headp = new_nodep;
                new_listp->headp->restp = temp_nodep;
                ++(old_listp->size);
            }
        cur_listp->headp = cur_listp->headp->restp;
        }
    }
    else{
        new_listp->headp = old_listp->headp;
    }
    return(new_listp->headp);
}

我称这个功能为像这样的功能;

listp->headp = place_first(listp, "Mustafa");
listp->headp = place_first(listp, "Mustafa");

My output like this: __Mustafa __Mustafa __Ataturk __Ali __Eisenhower __Kennedy Thanks for advance...

Answer 1

给

if ( old_listp->headp->name != name ){

您是在比较指针,而不是内容,您想要的

if (strcmp(old_listp->headp->name,name)){

其他检查也是如此。

Answer 2

感谢@dfan和@Daniel Fisher。

现在,我明白了我愚蠢的问题。

我对自己的职能做了一些改变。

那么,我做了什么.......

我为检查第一个节点添加了这一声明;

if ( strcmp(old_listp->headp->name, name) != 0 ){

并添加了用于检查其他节点的环环 :

for( cur_listp->headp = old_listp->headp;
                cur_listp->headp != NULL && strcmp(cur_listp->headp->name, name)!= 0;
                cur_listp->headp = cur_listp->headp->restp) { }

如果它返回无效,我的其他声明将排除否则旧节点将返回。

if ( cur_listp->headp == NULL){
...

完整功能(也许有人需要):

name_node_t *
place_first(name_list_t *old_listp, char name[11])
{
    name_list_t *new_listp, *cur_listp;
    name_node_t *new_nodep, *temp_nodep;
    temp_nodep = (name_node_t *)malloc(sizeof (name_node_t));
    new_listp = (name_list_t *)malloc(sizeof (name_list_t));
    cur_listp = (name_list_t *)malloc(sizeof (name_list_t));
    new_nodep = (name_node_t *)malloc(sizeof (name_node_t));
    cur_listp->headp = old_listp->headp;
    temp_nodep = old_listp->headp;
    new_listp = old_listp;
    if ( strcmp(old_listp->headp->name, name) != 0 ){
        for( cur_listp->headp = old_listp->headp;
                cur_listp->headp != NULL && strcmp(cur_listp->headp->name, name)!= 0;
                cur_listp->headp = cur_listp->headp->restp) { }
        if ( cur_listp->headp == NULL){
            if (old_listp->headp == NULL){
                strcpy(new_listp->headp->name, name);
                new_listp->headp->restp = NULL;
            }
            else {
                strcpy(new_nodep->name, name);
                new_nodep->restp = NULL;
                new_listp->headp = new_nodep;
                new_listp->headp->restp = temp_nodep;
                ++(old_listp->size);
            }
        }
        else
            new_listp->headp = old_listp->headp;
    }
    else{
        new_listp->headp = old_listp->headp;
    }
    return(new_listp->headp);
}

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