雙鏈表 - 不能寫迭代器給出節點正確的索引號

Question

我想實現拆分功能來學習C困難的方式雙鏈表，但要做到這一點，我需要每個節點都有他們的索引號作爲常規列表，數組都有。當我執行'推'功能，它將新節點添加到列表的末尾時，一切都很好，但是當我執行「不移位」時，將節點添加到列表頂部時，我遇到了使每個節點保持適當索引的問題編號從0開始。我希望在函數結束時創建一個迭代器，它遍歷每個節點並給出第一個0，第二個1 ..等我到現在爲止所做的，就是每個節點索引正在被更改爲0而不增加它。希望你們這些人能夠幫助我。

'list.h'  
#ifndef lcthw_List_h 
#define lcthw_List_h 

#include <stdlib.h> 

struct ListNode; 

// ListNode contains value, next, and prev struct. Each ListNode is another 
// chain in structure, they are linked 
typedef struct ListNode { 
    struct ListNode *next; 
    struct ListNode *prev; 
    void *value; 
    int track_num; 
} ListNode; 

// List is an guardian angel of ListNodes, it keeps tracking them by count 
// and knows which Node is first and last 
typedef struct List { 
    int count; 
    ListNode *first; 
    ListNode *last; 
} List; 


// Some standard functions for operate lists 
List *List_create(); 
void List_destroy(List *list); 
void List_clear(List *list); 
void List_clear_destroy(List *list); 

// These Macros return count, first and last element of the list 
#define List_count(A) ((A)->count) 
#define List_first(A) ((A)->first != NULL ? (A)->first->value : NULL) 
#define List_last(A) ((A)->last != NULL ? (A)->last->value : NULL) 

// List_push adds a new element to the end of the list 
void List_push(List *list, void *value); 
//List_pop takes the last element of the list and returns it 
void *List_pop(List *list); 

// Unshift adds element to the top of the list 
void List_unshift(List *list, void *value); 
// Shift retuns and remove first element from the list 
void *List_shift(List *list); 

// Removes list 
void *List_remove(List *list, ListNode *node); 

// This Macro is very useful, It Iterates through elements in the list 
#define LIST_FOREACH(L, S, M, V) ListNode *_node = NULL;\ 
    ListNode *V = NULL;\ 
    for(V = _node = L->S; _node != NULL; V = _node = _node->M) 

#endif 


'list.c sample with subject function' 
void List_unshift(List *list, void *value) 
{ 
    int i; 
    assert(list != NULL); 
    assert(list->count >= 0); 
    if(list->count > 0) { 
     assert(list->first != NULL); 
    } 

    ListNode *node = calloc(1, sizeof(ListNode)); 
    check_mem(node); 

    node->value = value; 

    if(list->first == NULL) { 
     list->first = node; 
     list->last = node; 


    } else { 
     node->next = list->first; 
     list->first->prev = node; 
     list->first = node; 


    } 

    list->count++; 

     LIST_FOREACH(list, first, next, cur) { 
      for(i = 0;i < list->count -1;i++) { 
       cur->track_num = i; 
      } 
     } 
error: 
    return; 
} 

Answer 1

「CUR」 是內部的for循環

for(i = 0;i < list->count -1;i++) { 
    cur->track_num = i; 
} 

不變和 「爲（I = 0;我<列表 - >計數-1」 是一個 '關閉由一個' 。說清單 - >計數等於1，清單 - >計數1是「0」和「for（i = 0; i < 0; i ++）」什麼都不做