如何在main函数中声明node类型的数组链并将其作为参数传递给其他函数？

Question

我正在尝试实现链接（哈希）。我不想全局声明

node *chain[size]

并希望它在 main 中声明它并通过函数将其作为参数传递。为此，我将代码修改为以下代码：

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

#define size 10

typedef struct hashNode
{
    int data;
    struct hashNode *next;
} node;

void insert(node *chain, int data)
{
    node *baby = (node*)malloc(sizeof(node));
    baby->data = data;
    baby->next = NULL;
    
    int key = data % size;
    if(chain[key]==NULL)
        chain[key] = baby;
    
    else
    {
        baby->next = chain[key];
        chain[key] = baby;
    }
}

int main()
{
    
    node *chain[size];
    
    for(int i=0; i<size; i++)
        chain[i] = NULL;
    
    insert(chain, 10);
    return 0;
}

代码给出了多个错误：

chaining.c: In function ‘insert’:
chaining.c:21:15: error: invalid operands to binary == (have ‘node’ {aka ‘struct hashNode’} and ‘void *’)
   21 |  if(chain[key]==NULL)
      |     ~~~~~~~~~~^~
      |          |
      |          node {aka struct hashNode}
chaining.c:22:16: error: incompatible types when assigning to type ‘node’ {aka ‘struct hashNode’} from type ‘node *’ {aka ‘struct hashNode *’}
   22 |   chain[key] = baby;
      |                ^~~~
chaining.c:27:16: error: incompatible types when assigning to type ‘struct hashNode *’ from type ‘node’ {aka ‘struct hashNode’}
   27 |   baby->next = chain[key];
      |                ^~~~~
chaining.c:28:16: error: incompatible types when assigning to type ‘node’ {aka ‘struct hashNode’} from type ‘node *’ {aka ‘struct hashNode *’}
   28 |   chain[key] = baby;

但是，如果我在结构之后全局声明数组链，则相同的代码可以完美运行：

typedef struct hashNode
{
    int data;
    struct hashNode *next;
} node;

node *chain[size];