CODEWITHSUNDEEP
c++classconstructorcopy
user13229658
user13229658

Reputation:

Copy constructor C++ does not copy pointer correctly

I created a copy constructor for my class. Does any one know why l1.ptr and l2.ptr give me the same address after compilation? I've done it many times and don't know where the error is.

#include<iostream>
#include<string>

using namespace std;

class numb
{
public:
    int* ptr;
    numb(int = 3);
    numb(const numb&);
    virtual ~numb();
};

numb::numb(int x)
{
    this->ptr = new int(x);
}

numb::numb(const numb& l1)
{
    this->ptr = new int(*(l1.ptr));

}

numb::~numb()
{    
}

int main()
{
    numb l1(5), l2;
    l1 = l2;
    cout << l1.ptr << endl;
    cout << l2.ptr << endl;
    system("Pause");
    return 0;
}

Upvotes: 0

Views: 491

Answers (2)

cigien
cigien

Reputation: 60228

In this snippet:

numb l1(5), l2;
l2 = l1;

the second line is not calling the copy constructor. Instead, it is calling the copy-assignment operator. Since you have not defined that, you get a shallow copy.

You can either use the copy constructor, like this:

numb l1(5);
numb l2(l1);

or define the operator= for your class:

numb& operator=(const numb&);  // do a deep copy

Upvotes: 4

bartgol
bartgol

Reputation: 1873

By not providing an assignment operator, you get the one that the compiler generates for you. The compiler is indeed "copying the pointer" (usually referred to as 'shallow copy').

If you need to copy the pointee (usually referred to as 'deep copy'), then you have to implement your own assignment operator.

Upvotes: 0

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