在c进行结构结构指针数组灵活的初始化 [英] initialization of flexible array of struct pointers in a struct in c

问题描述

我学习哈希表的数据结构，我想和指针的灵活长度阵列结构体的链接（链接列表件）散列表，这样，哈希表初始化将设置阵列是一个长度输入到初始化函数。

I'm learning hashtable data structures and I want to make a hashtable with a flexible length array of pointers to struct Link (linked list pieces), so that hashtable initialization will set the array to be a length input into the initialization function.

起初我得到的错误柔性阵列不是结构的终结。当其在端（如图所示）的程序崩溃（但它仍然编译）。这是我的code：

At first I was getting the error "flexible array not at the end of struct". When its at the end (as shown) the program crashes (but it still compiles). This is my code:

typedef struct Link{
    int key;
    char *name;
    struct Link *next;
} Link;

typedef struct HashTable{
    int numberOfEntries;
    int numberOfBuckets;
    Link *Table[];
} HashTable;

HashTable *hashtableInit(int size){
    HashTable *newHT = malloc(sizeof(HashTable));
        if (newHT != NULL){
            newHT->numberOfEntries = 0;
            newHT->numberOfBuckets = size;
            for (int i = 0; i < newHT->numberOfBuckets; i += 1){
                newHT->Table[i] = NULL;
            }
            return newHT;
        } else {
            printf("Error in memory allocation.\n");
            fflush(stdout);
            return NULL;
        }
    }
}

它的工作原理，如果我数组常量，输入相同的值设置到初始化函数：

It works if I set the array to a constant and input the same value into the init function:

#define SIZE 11

typedef struct Link{
    int key;
    char *name;
    struct Link *next;
} Link;

typedef struct HashTable{
    Link *Table[SIZE];        
    int numberOfEntries;
    int numberOfBuckets; 
} HashTable;

HashTable *hashtableInit(int size){ // works if SIZE is passed into function as size parameter
    HashTable *newHT = malloc(sizeof(HashTable));
        if (newHT != NULL){
            newHT->numberOfEntries = 0;
            newHT->numberOfBuckets = size;
            for (int i = 0; i < newHT->numberOfBuckets; i += 1){
                newHT->Table[i] = NULL;
            }
            return newHT;
        } else {
            printf("Error in memory allocation.\n");
            fflush(stdout);
            return NULL;
        }
    }
}

第二code座完美的作品。任何见解将大大AP preciated。谢谢你的时间。
克里斯

The second code block works perfectly. Any insights would be greatly appreciated. Thanks for your time. Chris

推荐答案

您应该分配内存

HashTable *newHT = malloc(sizeof *newHT + size * sizeof newHT->Table[0]);

这篇关于在c进行结构结构指针数组灵活的初始化的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！