What happens to a unique pointer type after using std::move?

[brianbathorycpp]

If do something like this:

int main()
{
    std::unique_ptr<int> mInt = std::make_unique<int>();
    
        // Move contents of mInt into mInt2
    std::unique_ptr<int> mInt2 = std::move( mInt );
    return 0;
}

What ends up happening to the first unique pointer, mInt? Does it get destroyed or is it left as a nullptr?

Thanks for taking the time!..

[Ganado]

In general, when you call the move assignment operator, the first object is left in an unspecified (but still destructible) state and should not be used.

However, the above is only a general rule of thumb. The standard library does make some guarantees for certain standard library classes, including unique_ptr.
See: https://stackoverflow.com/questions/36071220/what-happens-to-unique-ptr-after-stdmove

mInt.get() is guaranteed to be null after the move assignment.

[brianbathorycpp]

Oh I understand...

mInt.get() is guaranteed to be null after the move assignment

Which means I could return the same contents back to it, right?

if( mInt.get() == nullptr )
{
    mInt = std::move( mInt2 );
}

Maybe something like that?

[Ganado]

Yes, that is legal. Maybe somebody else might have more useful "best practices" advice than me.

[brianbathorycpp]

Yep! It works perfectly, thank you!..

[seeplus]

Note that the code would be simpler if you used auto - cuts down the typing and the chance of typo mistakes!

#include <memory>

int main()
{
	auto mInt {std::make_unique<int>()};

	// Move contents of mInt into mInt2
	auto mInt2 {std::move(mInt)};
}