Dino Dino - 3 months ago 10
C++ Question

Object not deleted before new is assigned

I'm kind of confused, because I was sure this should work different. Take a look at this code example:

#include <iostream>
#include <string>

using namespace std;

class base
{
public:
virtual ~base() = default;
};

class derived : public base
{
private:
int a = 0;
int *b = nullptr;
std::string lol;

public:
derived(std::string s) : b(new int(6)), lol{s} { cout << "ctor " << lol << endl; }
derived(derived const& d) : lol{d.lol + " copy"} {cout << "copy " << lol << endl; }

virtual ~derived() { cout << "dtor " << lol << endl; delete b; }

virtual void superFunction() { cout << "OMG " << lol << endl; }
};

int main()
{
derived a("a");
derived b("b");
a = b;
}


And the program output with all optimizations off is:

ctor a
ctor b
dtor b
dtor b


I was sure that in this case compiler should generate code that deletes object
a
and uses copy constructor to create new object. Instead it uses
operator=
that it implicitly declares.

Can someone explain why? Or point me to C++ standard.

Thanks.

Answer

When you write a = b;, compiler calls assignment operator, which will be automatically generated if not present in the code and not marked as deleted. Copy constructor is used only if you try to initialize a new object from another object like this:

derived a("a");
derived b = a;

Also, your code crashes before main returns as it tries to delete b, which points to the same memory from a and from b after a = b; default-assignment.

If you want to delete a with derived destructor after a = b; execution, all you need is copy-and-swap idiom. Here you may find a great answer on how to do that in legacy and modern C++. Proper implementation of rule-of-four from that answer will perfectly fit the DRY principle and help you to avoid memory issues. Note the fabulous trick with passing parameter to operator= by value, which makes compiler select the appropriate constructor (copy or move) and allows you to write only four methods instead of all five of them.

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