Question

我尝试创建一个空列表数组，然后从main（）和Graph（）构造函数中打印这些列表的位置。但他们不匹配！请帮忙！

#include<iostream>
#include<list> // use list in STL

using namespace std;

struct Edge {

  int v, w;

  Edge(int v = -1, int w = 0) : v(v), w(w){}

};

class Graph {

  private:

    const int V; //number of vertices

    int E; //number of edges

    list<Edge>* adj[];//adjacency list

  public:

    Graph(int N):V(N){ //create a graph with no edges

      E = 0; list<Edge>* adj[V];

      for(int i = 0; i < N; i++){

        adj[i] = new list<Edge>();

        cout << "at " << i <<" address "<< adj[i] <<"\n";

      }

    }

   friend ostream& operator<<(ostream& out, const Graph& G);

};

ostream& operator<<(ostream& out, const Graph& G){

  out << "Vertices: "<< G.V <<" and Edges: "<< G.E <<"\n" << "The adjacency list:\n";

  for(int i = 0; i < G.V; i++){

    out << "vertex "<< i << ":"<< G.adj[i] <<" -> zero\n";

  }

  return out;

}

int main() {

  Graph* G = new Graph(50);

  cout << *G;

}

Answer 1

Johnny Mopp，flatmouse

感谢。

更改为以下代码。它奏效了。

Graph（int N）：V（N）{

  E = 0;

  for(int i = 0; i < N; i++){

    adj[i] = new list<Edge>(); 

   }

}

~Graph(){

    for(int i = 0; i < V; i++)

      delete adj[i];

}

示例输出：

在0地址0xec3c40

在1地址0xec4070

在2地址0xec4090

在3地址0xec40b0

//

顶点0：0xec3c40 - ＆gt;零

顶点1：0xec4070 - ＆gt;零

顶点2：0xec4090 - ＆gt;零

顶点3：0xec40b0 - ＆gt;零

列表数组 - 构造函数内和main（）内的地址不匹配

1 个答案: