在结构时间3内访问结构？

Question

我在C中有一个作业，并且在访问结构中的不同成员时遇到麻烦（某些级别很深）。我了解基本原理，但在某些地方会有所遗漏。我有3个结构，最上面的一个包含第二个的数组，而后者又包含第三个的数组。我当前的问题是正确使用malloc。这是我的一些代码。我将不胜感激任何信息或提示，因为我还有很长的路要走，而且正如您所看到的，结构有点复杂。

.h文件

typedef struct user {
    char* userID;
    int wallet;
    bitCoinList userBC; //Also a list
    senderTransList userSendList; //Yes it has lists too..
    receiverTransList userReceiveList;
}user;

typedef struct bucket {
    struct bucket* next;
    user** users;
}bucket;

typedef struct hashtable {
    unsigned int hashSize;
    unsigned int bucketSize;
    bucket** buckets;
}hashtable;

这是我创建和初始化哈希表的功能。当我尝试使用HT->buckets->users访问用户时（请求成员用户不是结构体或联合体），会收到错误消息

.c文件

// Creation and Initialization of HashTable
hashtable* createInit(unsigned int HTSize,unsigned int buckSize){

    hashtable* HT = (hashtable*)malloc(sizeof(hashtable));
    if(HT==NULL) {
        printf("Error in hashtable memory allocation... \n");
        return NULL;
    }

    HT->hashSize=HTSize;
    HT->bucketSize=buckSize;

    HT->buckets = malloc(HTSize * sizeof(HT->buckets));
    if(HT->buckets==NULL) {
        printf("Error in Buckets memory allocation... \n");
        return NULL;
    }
    HT->buckets->users = malloc(buckSize * sizeof(HT->buckets->users));
    if(HT->buckets->users==NULL) {
        printf("Error in Users memory allocation... \n");
        return NULL;
    }
    for(int i=0; i <HTSize; i++){
        HT->buckets[i] = malloc(sizeof(bucket));
        HT->buckets[i]->next = NULL;
        if(HT->buckets[i]==NULL) {
            printf("Error in Bucket %d memory allocation... \n",i);
            return NULL;
        }

        for(int j=0; j <buckSize; j++){
            HT->buckets[i]->users[j] = malloc(sizeof(user));
            if(HT->buckets[i]==NULL) {
                printf("Error in User %d memory allocation... \n",i);
                return NULL;
            }
        }
    }
    return HT;
}

Answer 1

因为存储桶是指向指针类型的指针，所以您需要：

(*(HT-> buckets)) ->users = ....

或

HT-> buckets[0] ->users = ....   // or any other index depending of the program logic

或（对于第n个指针）

(*(HT-> buckets + n)) ->users = ....

或

HT-> buckets[n] ->users = ....   // or any other index depending of the program logic

这只是语法答案，我不分析程序逻辑

Answer 2

至少一个问题：大小分配错误。

分配给HT->buckets所指向的数据的大小，而不是指针的大小。

避免错误。以下成语很容易编码，检查和维护。

ptr = malloc(sizeof *ptr * n);

---

// HT->buckets = malloc(HTSize * sizeof(HT->buckets));
HT->buckets = malloc(HTSize * sizeof *(HT->buckets));

// HT->buckets->users = malloc(buckSize * sizeof(HT->buckets->users));
HT->buckets->users = malloc(buckSize * sizeof *(HT->buckets->users));

// HT->buckets[i] = malloc(sizeof(bucket));
HT->buckets[i] = malloc(sizeof *(HT->buckets[i]));

// HT->buckets[i]->users[j] = malloc(sizeof(user));
HT->buckets[i]->users[j] = malloc(sizeof *(HT->buckets[i]->users[j]));