如何在c的主函数中声明结构变量？

Question

我是 c 编程的新手，我的代码正在运行，但我的问题是，如果我在 main 函数中声明 struct node *a,*b;，如何将 a 和 b 传递给 void create()。为什么它不起作用，有人可以帮我理解吗？

#include<stdio.h>
#include<conio.h>
#include<malloc.h>

struct node
{
int d;
struct node *next;
}*start=NULL;struct node *a,*b; //move this part to main function -> struct node *a,*b; but its not working

void create()
{
    a=(struct node *)malloc(sizeof(struct node));
    printf("Enter the data : ");
    scanf("%d",&a->d);
    a->next=NULL;

    if(start==NULL)
    {
        start=a;
        b=a;
    }

    else
    {
        b->next=a;
        b=a;
    }
}

void display()
{

    struct node *a;
    printf("\nThe Linked List : ");
    a=start;

    while(a!=NULL)
    {
        printf("%d--->",a->d);
        a=a->next;
    }
    printf("NULL\n");
}



void main()
{

    char ch;

    do
    {
        create();   
        printf("Do you want to create another : ");
        ch=getche();
    }

    while(ch!='n');

    display();

}
void freenodes()
{
    struct node *a;
    a = start;

    while(a != NULL)
    {
      struct node *freenode = a ;
      a = a->next;
      free(freenode) ;
    }
}

Answer 1

void main()
{
  struct node name_of_varialbe;//this is struct variable declaration
  ...
}    

Answer 2

只需将void create()的原型改成void create(struct node *,struct node *)，这样调用：

char ch;

do
{
    struct node *a_main=(struct node *)malloc(sizeof(struct node));
    struct node *b_main=(struct node *)malloc(sizeof(struct node));
    create(a_main,b_main);   
    printf("Do you want to create another : ");
    ch=getche();
}while(ch!='n');

void create(struct node *a,struct node *b)
{
    //do your stuff
}