取消引用void *指针，结构数组

Question

I am trying to change a void * pointer into an array of structs. The point is to have a global visible pointer initialized to NULL, that when main starts, will be turned to an array of structs Here is a minimal example.

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

void * hashtable;

struct bucket {
    int a;
    int b;
};

int main (void)
{
    hashtable = (struct bucket)malloc(6*sizeof(struct bucket));
    int i ;
    //for(i=0;i<6;i++)
    //  hashtable[i] =  malloc(sizeof(struct bucket));

    *(struct bucket)hashtable[0]->a = 12;

    return 0;
}

The errors I get are:

test.c:16:52: error: conversion to non-scalar type requested
  hashtable = (struct bucket)malloc(6*sizeof(struct bucket));
                                                ^
test.c:21:27: warning: dereferencing ‘void *’ pointer
  *(struct bucket)hashtable[0]->a = 12;
                       ^
test.c:21:27: error: void value not ignored as it ought to be

Answer 1

As mentioned already, you shouldn't be casting the result of malloc() here:

hashtable = (struct bucket)malloc(6*sizeof(struct bucket));

Trim it down to:

hashtable = malloc(6*sizeof(struct bucket));

---

As for your assignment:

*(struct bucket)hashtable[0]->a = 12;

1. Your program first tries to compute hashtable[0]->a , but that gives you that compiler warning since you haven't cast hashtable yet ( [] and -> have higher precedence than casting).

2. You need to cast to struct bucket * since hashtable is a pointer.

3. You attempt to dereference hashtable too many times; you don't need the extra * and -> operators.

With this in mind, you can access a of hashtable 's first element as follows:

((struct bucket *)hashtable)[0].a = 12;

Of course, you might have an easier time defining struct bucket * hashtable; to avoid casting hashtable every time you want to work it.