为什么我的代码在我写“G->Adj[i]”但在图形实现（邻接表）中写“G->Adj+i”时不起作用。？

Question

#include <stdio.h>
#include <conio.h>
#include <stdlib.h>
struct Graph* GraphList();
struct Listnode
{
     int vertex;
     struct Listnode*next;
};
struct Graph
{
  int V;
  int E;
  struct Listnode *Adj;
};
void main()
{
    struct Graph *G=NULL;
    int i;
    struct Listnode *temp=NULL;
    printf("Program to Implement graph using Adjacency List\n");
    G=GraphList();
     for(i=0;i<G->V;i++)
     {
         temp=G->Adj+i;
         while((temp->next)!=(G->Adj+i))
            {
                printf("%d-->",temp->vertex);
                temp=temp->next;
            }
        printf("%d",temp->vertex);
        printf("\n");
     }
}
struct Graph* GraphList()
{
     int i,j,x,y;
     struct Listnode *t,*temp;
      struct Graph *G;
      G=(struct Graph*)malloc(sizeof(struct Graph));
      printf("Enter the no.of vertices and Edges respectively\n");
      scanf("%d %d",&G->V,&G->E);
      G->Adj=(struct Listnode*)malloc((sizeof(struct Listnode))*G->V);  //Undirected Graph
      for(i=0;i<G->V;i++)
      {
          (G->Adj+i)->vertex=i+1;
       (G->Adj+i)->next= (G->Adj+i);
      }
      for(i=0;i<G->V;i++)
      {
          t= (G->Adj+i);
          printf("Enter number of neigbouring nodes to node-%d  ",i+1);
          scanf("%d",&x);
          for(j=0;j<x;j++)
          {
          printf("Enter the vertex number-%d that is neighbour to node -%d  ",j+1,i+1);
              scanf("%d",&y);
              temp=(struct Listnode*)malloc(sizeof(struct Listnode));
              temp->vertex=y;
              temp->next= G->Adj+i;
              while(t->next!= (G->Adj+i))
                     t=t->next;
              t->next=temp;
          }
      }
      return G;
}

上面的代码工作正常，但是当我用“G->Adj[i]”替换“G->Adj+i”时它不起作用。我创建了“struct Listnode”类型的数组（比如说size-5）并将其存储在“G->Adj”中，我觉得使用“G->Adj[i]”是公平的，但我没有不明白为什么会弹出错误信息：

"D:\DataStructures by SS\Graph Link (standard).c|60|error: incompatible types when assigning to type 'struct Listnode *' from type 'struct Listnode'| ".

请解释一下。

Answer 1

G->Adj+i 等价于&G->Adj[i]。要撤消 [i] 的取消引用，需要前导 & 号。