这会产生一个不兼容的警告:
#include <stdlib.h>
#include <stdio.h>
typedef struct
{
int key;
int data;
struct htData_* next;
struct htData_* prev;
}htData_;
typedef struct
{
int num_entries;
struct htData_** entries;
}ht_;
ht_* new_ht(int num_entries);
int ht_add(ht_* ht_p, int key, int data);
int main()
{
int num_entries = 20;
//crate a hash table and corresponding reference
ht_* ht_p = new_ht(num_entries);
//add data to the hash table
int key = 1305;
ht_add(ht_p,key%num_entries,20);
return 0;
}
ht_* new_ht(int num_entries)
{
ht_ *ht_p;
ht_ ht;
ht.num_entries = num_entries;
ht_p = &ht;
//create an array of htData
htData_ *htDataArray;
htDataArray = (htData_*) malloc(num_entries * sizeof(htData_));
//point to the pointer that points to the first element in the array
ht.entries = &htDataArray; // WARNING HERE!!!!!!!!!!!!!!!!
return ht_p;
}
我正在尝试将
**ptr
复制到包含**ptr
的struct
中。
更新: 我简化的代码不准确,所以我发布了实际代码。
htData_ array[20] = htDataArray;
的目的是什么? - Mark Elliot