Why is this function returning 1?

Question

Although the control isn't going to the if condition in isCyclic() function but still it is returning 1. When I uncomment the return 0 statement, it is indeed returning 0.

So, does an int function return 1 by default? I don't think so. It returns garbage value. But then why it is returning 1 here!!!

#include <bits/stdc++.h>

using namespace std;

class edge{
public:
    int src,des;
};

class graph{
public:
    int v,e;
    edge *edges;
    int edges_counter;
    graph(int v, int e){
        this->v = v;
        this->e = e;
        edges = new edge[e];
        edges_counter = 0;
    }
    void addedge(int src, int des){
        edges[edges_counter].src = src;
        edges[edges_counter].des = des;
        edges_counter++;

    }

    int find(int parent[], int i){
        if(parent[i]==-1)
            return i;
        return find(parent, parent[i]);
    }

    void Union(int parent[], int x, int y){
        parent[x] = y;
    }

    int isCyclic(){
        int parent[v];
        for(int j=0; j<v ;j++) parent[j] = -1;

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

            int first = find(parent, edges[i].src);
            int second = find(parent, edges[i].des);

            if(first==second){
                cout <<"Hell";
                return 1;
            }

            Union(parent, first, second);

        }
        //return 0; //here
    }
};

int main(){
    graph g(3,2);
    g.addedge(0, 1);
    g.addedge(0, 2);
    cout << g.isCyclic() << endl;
    if(g.isCyclic()==1) cout << "Grpah contains cycle\n";
    else cout << "Grpah doesn't contain cycle\n";
}

You should not use `#include `, use the `#include`s you need. — mch, May 10 '17 at 07:21

score 2 · Answer 1 · edited May 23 '17 at 12:26

2

And isn't 1 a perfect "garbage value"?

The C++11 draft states in § 6.6.3/2:

Flowing off the end of a function [...] results in undefined behavior in a value-returning function.

"undefined behavior" basically means "anything can happen". If you have a function returning "int", but you don't return, it'll return whatever is currently laying around in memory. That could be 0, 1, 12345, or any other garbage value you can think of.

You should add a return statement to every possible code flow where your function could end. You can use the parameter -Wreturn-type to make the compiler warn you about these problems in the future.

edited May 23 '17 at 12:26

Community

1
1

answered May 10 '17 at 07:11

Florian Bach

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but it is returning 1 every time on every PC!! – Ashutosh Maheshwari May 10 '17 at 07:20
So what? It is allowed to do that. "Undefined" does not mean "random". It *can* do so, but it doesn't have to. If you want a specific outcome, define it with "return" and add any return value you want. – Florian Bach May 10 '17 at 07:24

eerorika · Accepted Answer · 2017-05-10T07:32:37.943

So, does an int function return 1 by default?

No. There is no default return value (except for main, which returns 0 by default.)

It returns garbage value.

Correct. More correctly: The behaviour of the program is undefined.

But then why it is returning 1 here!!!

Because the behaviour is undefined.

I think you misunderstand what "garbage" value means. Garbage can have any value. 1 is within the set "any value".

but it is returning 1 every time on every PC!!

Because the behaviour is undefined. The standard does not guarantee that the return value be other than 1 on any occasion. Considering that, why would you expect the function to not return 1? Of course, it doesn't guarantee that the return value be 1 either, so expecting that would be equally silly. Any expectation would be folly when you know that the behaviour is undefined.

Yeah I agree but it is returning 1 every time on every PC!! – Ashutosh Maheshwari May 10 '17 at 07:21 — Ashutosh Maheshwari, May 10 '17 at 07:21

Why is this function returning 1?

2 Answers2