This code is giving me runtime error. Here, first I want to take a integer input which determine the number (n) of nodes in the graph. Then for n-1 time I want take two integer input which will give the adjacency information.

#include <set>
#include <map>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <cassert>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>

#define MAXEDGE 1000


using namespace std;


typedef struct{
    int linked_to;
}edg;

typedef struct{
    edg edge[MAXEDGE];
    bool visited[MAXEDGE];
    int total_nodes;
}graph;
	
int main() {
    
        int u,v,n = 0;
        graph *g;
        
        scanf("%d",&g->total_nodes);
       
        for(int i = 0; i<g->total_nodes; i++){
                scanf("%d %d",&u,&v);
                g->edge[u].linked_to = v;
        }        
    
    return 0;
}

Where did you allocate the memory

Sorry, I don't understand your question. Why I allocate memory here?

g is pointer to an object of type graph,
you have to create the object, by either malloc or new
and then point g to the address

graph *g; creates a pointer (to memory) but you fail to provide any memory to point to. You can do one of three things:

  • Create an object instead of a pointer ( graph g; )
  • Create an object and assign the pointer to that objects address ( graph g, *gp = &g; )
  • Allocate the memory yourself ( graph * g = new graph; )

Edited 4 Years Ago by L7Sqr: n/a

graph *g; creates a pointer (to memory) but you fail to provide any memory to point to. You can do one of three things:

  • Create an object instead of a pointer ( graph g; )
  • Create an object and assign the pointer to that objects address ( graph g, *gp = &g; )
  • Allocate the memory yourself ( graph * g = new graph; )

Thank You. 2nd one is working for me. But what about 1st one? How can I assign value for the object(g) using 1st option?

Would you plz make me clear about how 2nd one is working? Is it like this...
For example, pointer gp is first pointing the object g which is a graph type object. then the value of total_nodes is assigned through the pointer to the object.

When you have syntax like g->total_nodes it means that you access the object pointed to by "g" and then access its data member called total_nodes .

When you write it as g.total_nodes it means that you access the data member called total_nodes that belongs to the object "g".

So, if you want the first version to work, i.e. the version with Graph g; , then you have to replace all the g-> with g. , because now, "g" is an object, and not a pointer to an object (like in your original code).

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