创建具有其他列表元素的列表(结构相同) [英] Creating a list with elements other lists (same struct)

问题描述

假设我有3个列表:list1，list2，list3.

Lets say i have 3 lists : list1,list2,list3.

这些列表的每个元素的 struct :

Their struct of each element of those lists :

struct node {
    char value[20] ;
    struct node * next ;    
    int occurs;
} ;

typedef struct node Node;
typedef Node * List;

但是我不认为这很重要.

but i dont think it matters.

我想创建一个新列表，但是它的每个元素都必须是这3个列表中的每一个.对此，我的新 struct 是(正确的):

I want to create a new list but every element of it has to be each of those 3 lists.My new struct for this matter is that(its correct):

typedef struct listoflists{
    List list;
    struct listoflists*next;
}Nested; 

我创建列表的新功能:

void createlistoflists(Nested **LIST,List list1,List list2,List list3){
     if (*LIST==NULL){
         *LIST=listab;
     }
     else

所以我不确定乞求是否正确，但是我如何填写(并正确)以获取列表列表?

So im not sure if the begging is even correct but how will i fill it(and correct) in order to achieve the list of lists?

推荐答案

Nested* Nested_create(List list) {
    Nested* new = malloc(sizeof(Nested));
    new->list = list;
    new->next = NULL;
    return new;
}

void Nested_add(Nested** proot, Nested* node) {
    if (*proot == NULL) {
        *proot = node;
    } else {
        Nested* cur = *proot;
        while (cur->next)
            cur = cur->next;
        cur->next = node;
    }
}

void createlistoflists(Nested **LIST, List list1, List list2, List list3) {
    Nested_add(LIST, Nested_create(list1));
    Nested_add(LIST, Nested_create(list2));
    Nested_add(LIST, Nested_create(list3));
}

在一个函数中:

void createlistoflists(Nested **LIST, List list1, List list2, List list3) {
    List lists[] = {list1, list2, list3};
    for (List* it = lists; it < lists + 3; ++it) {
        Nested* node = malloc(sizeof(Nested));
        node->list = *it;
        node->next = NULL;
        *LIST = node;
        LIST = &(*LIST)->next;
    }
}

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