Why std::move is needed? [duplicate]

Question

I read a beautiful article on the move semantics in C++11. This article is written in a very intuitive way. The example class in the article is given below.

class ArrayWrapper { public: // default constructor produces a moderately sized array ArrayWrapper () : _p_vals( new int[ 64 ] ) , _metadata( 64, "ArrayWrapper" ) {} ArrayWrapper (int n) : _p_vals( new int[ n ] ) , _metadata( n, "ArrayWrapper" ) {} // move constructor ArrayWrapper (ArrayWrapper&& other) : _p_vals( other._p_vals ) , _metadata( other._metadata ) { other._p_vals = NULL; } // copy constructor ArrayWrapper (const ArrayWrapper& other) : _p_vals( new int[ other._metadata.getSize() ] ) , _metadata( other._metadata ) { for ( int i = 0; i < _metadata.getSize(); ++i ) { _p_vals[ i ] = other._p_vals[ i ]; } } ~ArrayWrapper () { delete [] _p_vals; } private: int *_p_vals; MetaData _metadata; };

Clearly in the above move constructor implementation, the movement doesn't happen for the the embedded element _metadata. To facilitate this the trick is to use the std::move() method like this.

ArrayWrapper (ArrayWrapper&& other) : _p_vals( other._p_vals ) , _metadata( std::move( other._metadata ) ) { other._p_vals = NULL; }

So far, so good.

The standard says:

§5 (C++11 §5[expr]/6):

[ Note: An expression is an xvalue if it is:

the result of calling a function, whether implicitly or explicitly, whose return type is an rvalue reference to object type,

a cast to an rvalue reference to object type,

a class member access expression designating a non-static data member of non-reference type in which the object expression is an xvalue, or

a .* pointer-to-member expression in which the first operand is an xvalue and the second operand is a pointer to data member.

My question:

Now, the variable other in the move constructor is an xvalue (am I right?). Then according to the last rule above, other._metadata should also be an xvalue. And hence the compiler can implicitely use the move constructor of _metadata's class. So, no need to std::move here.

What am I missing?

It's a good idea to avoid leading underscores to prevent accidentally using a reserved name. — Mark B
– Mark B, Commented Jul 30, 2012 at 15:31
Good habit to get into. In this example, none start with capitals, so you're ok, but you can easily forget not to do that. — Qaz
– Qaz, Commented Jul 30, 2012 at 15:41
@MarkB : It is not my code. I just reproduced it from the refered article. — PermanentGuest
– PermanentGuest, Commented Jul 30, 2012 at 16:09

David Rodríguez - dribeas · Accepted Answer · 2012-07-30 15:37:17Z

Your assumption is not really true. The argument to the constructor is an xvalue, which allows the rvalue-reference to be bound, but once the rvalue-reference is bound, inside the constructor, it is no longer an xvalue but an lvalue. Conceptually, the object at the call place is expiring, but inside the constructor and until it completes it is no longer expiring, as it can be used later within the constructor block.

ArrayWrapper f(); ArrayWrapper r = f(); // [1]

In [1], the expression f() refers to a temporary, that will expire after the call to the constructor, so it can be bound by an rvalue-reference.

ArrayWrapper (ArrayWrapper&& other) : _p_vals( other._p_vals ) , _metadata( other._metadata ) // [2] { other._p_vals = NULL; std::cout << other._metadata << "\n"; // [3] }

Inside the constructor, other is not expiring, it will be alive for each and every instruction of the constructor. If the compiler allowed moving in [2], than a potential further use of the variable in [3] would be invalid. You have to explicitly tell the compiler that you want the value to expire now.

OK, I think I get it. So, this means, f() in [1] is an xvalue, but, 'other' in the function parameter is not an xvalue, right?
@PermanentGuest: right, an rvalue-reference (once bound) is an lvalue.

R. Martinho Fernandes · Accepted Answer · 2012-07-30 15:32:25Z

other is an lvalue, because it's a variable. Named references are lvalues, regardless of what kind of reference they are.

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