compiler defined move constructor with destructor

Question

As per the cpp reference cpp-ref, compiler does not generate a default move constructor if we have a user defined destructor.

Code snippet:

 class General
{
public:
    ~General();
    General();
    void testInitList();
};

int main(int argc, char **argv) {
    General b(std::move(General()));
    General g = std::move(b);
    g.testInitList();
    return 0;
}

The code compiles implying that the compiler generated a default move constructor. The code was compiled using gcc version 5.4.0.

Could someone explain why the compiler generated a move constructor and move assignment operator in this case despite have a destructor?

Best, Rahul

First, it compiled. Second, i had print statements in constructor and destructor. Only one constructor statement was printed where as multiple destructor statements were printed. — RGs
– RGs, Commented Jun 16, 2019 at 1:28
You don't define a copy constructor, that's why you don't have enough printings. — 3CEZVQ
– 3CEZVQ, Commented Jun 16, 2019 at 1:30
Does this answer your question? When does compiler not generate move constructor and move assignment? — JaMiT
– JaMiT, Commented Jun 26, 2023 at 5:29

Nikos C. · Accepted Answer · 2019-06-16 01:28:18Z

2

When there's no move constructor or assignment operator, no move is performed. std::move doesn't perform the move. It just casts its argument to indicate that a move may be performed if possible. If not possible, then there's no move and using std::move does nothing.

answered Jun 16, 2019 at 1:28

Nikos C.

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3 Comments

RGs Over a year ago

I had print statements in constructor and destructor. Only one constructor statement was printed where as multiple destructor statements were printed. It implies that the object was indeed moved. Output - Creating General Destroying General Destroying General Destroying General

RGs Over a year ago

Understood where I was making the mistake. Thank you so much Nikos.

Nikos C. Over a year ago

@RGs Your code calls the default constructor and the implicitly defined copy constructor. So yes, only one will be printed. Define a copy constructor and print something there.

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