CODEWITHSUNDEEP
c++constructorcopy-constructorexplicit
M.B.
M.B.

Reputation: 37

Uncertain behaviour of copy constructor in C++

As a student , I was testing out various ways of invoking the copy constructor . I came across 3 of them :

Test t2 = t1 // 1 ; 
Test t3(t1) // 2 ; 
Test t4 ;  // 3 ;
t4 = t1 ;

The explicitly defined copy constructor adds 10 to each of the data members of t1 . The first 2 work fine and call the explicit one . However the 3rd approach calls the implicit copy constructor i.e. there is no addition of 10 taking place .

My code :

#include <iostream>
using namespace std ; 
class Test
{
    int m,n ; 
    public: 
    Test() ; 
    Test(int,int) ; 
    ~Test() ; 
    Test(Test&) ; 
    void showData() ; 
}; 
inline Test::Test(Test& t)
{
    cout<<"called"<<endl ;
    m = t.m + 10 ; 
    n = t.n + 10 ; 
}
inline Test::Test(){}
inline Test::Test(int m,int n)
{
    Test::m = m ; 
    Test::n = n ; 
}
inline Test::~Test(){}
inline void Test::showData()
{
    cout<<m<<" "<<n<<endl ; 
}
int main()
{
    Test t1(3,4) ; 
    t1.showData() ; 
    Test t2(t1) ; 
    t2.showData() ;
    Test t3 = t1 ;
    t3.showData() ;
    Test t4 ; //calls the implicit one 
    t4 = t1 ; 
    t4.showData() ;
    return 0 ; 
}

Output :

3 4
called
13 14
called
13 14
3 4

I tried overloading the assignment(=) operator to manually solve the problem with the 3rd approach which did work , but I want to know why the 3rd approach calls the implicit copy constructor even though an definition has been provided explicitly .

Upvotes: 0

Views: 83

Answers (1)

Ahmed AEK
Ahmed AEK

Reputation: 18090

These invoke a copy constructor:

Test t2 = t1;  // calls Test::Test(const Test& t)
Test t3(t1);   // also calls Test::Test(const Test& t)

While this invokes a copy assignment operator:

Test t4 ;  // calls Test() constructor
t4 = t1 ;  // calls Test& operator=(const Test& t)

Both the copy constructor and the copy assignment operator will have a default implementation created by the compiler if you don't define one yourself.

Having the const Test& as a parameter allows you to invoke it on const objects. Your code without the const won't compile with const Test t1;.

Upvotes: 5

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