CODEWITHSUNDEEP
c++
Mohsin Mushtaq
Mohsin Mushtaq

Reputation: 93

Calling the default constructor

This program gives the correct output, but I can't understand how. How is the default constructor called at the time of the declaration of the object?

#include <iostream>
using namespace std;
class GuessMe {
private:   
    int *p;
public:
    GuessMe(int x=0) 
    {
        p = new int;
    }
    int GetX() 
    {
        return *p;
    }
    void SetX(int x) 
     {
        *p = x;
     }
    ~GuessMe() 
     {
        delete p;
     }
};

 int main() {
    GuessMe g1;
    g1.SetX(10);
    GuessMe g2(g1);
    cout << g2.GetX() << endl;
    return 0;
}

Upvotes: 0

Views: 82

Answers (1)

juanchopanza
juanchopanza

Reputation: 227390

This constructor has a default parameter:

GuessMe(int x=0)

That means that when a GuessMe is default constructed, it is as if it had been called with an argument with value 0. Note that the constructor parameter isn't used for anything in your code. Also note that p is set to point to an uninitialized integer here:

p = new int;

so calling GetX() before calling SetX() would yield undefined behaviour. Presumable you want to use the value of x to set p:

GuessMe(int x=0) 
{
    p = new int(x);
}

or, using initialization instead of assignment,

GuessMe(int x=0) : p(new int(x))
{
}

Also, read up on the rule of three to avoid double-deletes. And then learn to code without raw pointers to dynamically allocated objects.

Upvotes: 5

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