Creating a temporary of decltype

Question

I have an object of some type, for example, std::vector<int> v;
Now, say, I want to verify that v releases all its internal memory.
Prior to the C++11 shrink_to_fit() method, the recommended/guaranteed way to do this is to swap() with an empty std::vector<> of the same type.

However, I don't want to specify the type of the object. I can use decltype to specify the type, so I'd like to write something like this:

std::vector<int> v;
// use v....
v.swap(decltype(v)()); // Create a temporary of same type as v and swap with it.
                  ^^

However, the code above does not work. I cannot seem to create a temporary of type decltype(v) with an empty ctor (in this case).

Is there some other syntax for creating such a temporary?

Answer 1

The issue is that swap takes an lvalue reference: You cannot pass a temporary to swap. Instead you should switch it around so that you call the temporary's swap member:

decltype(v)().swap(v);

Of course C++11 introduced the shrink_to_fit() member so that the swap trick is no longer necessary.

Answer 2

You can't bind temporary rvalue to a non-const lvalue reference, and so you can't pass it as the argument to swap. However, you can call a member function on a temporary, so this will work:

decltype(v)().swap(v);

In C++11, it would be clearer to move from the temporary:

v = decltype(v)();

or use shrink_to_fit

v.clear();
v.shrink_to_fit();

(Note that, if you don't have C++11, then you don't have decltype, so the question is moot).

Answer 3

You may do (C++11):

v.clear();
v.shrink_to_fit();

or :

{
    decltype(v) tmp;

    v.swap(tmp);
}

vector::swap takes argument by reference. So temporary cannot be used here.

Answer 4

To clarify, decltype(v)() is the valid syntax for creating an unnamed temporary.
The source of the error, as the other answers explain, was that swap() does not work with the temporary returned by decltype(v)().