The function std::move()
is defined as
template<typename T>
typename std::remove_reference<T>::type&& move(T && t)
{
return static_cast<typename std::remove_reference<T>::type&&>( t );
}
There are four places where I can imagine the move constructor to be called:
std::move()
function itself but possibly at the place where the returned reference ultimately arrives. I would bet for number 4, but I'm not 100% sure, so please explain your answer.
A move constructor is called: when an object initializer is std::move(something) when an object initializer is std::forward<T>(something) and T is not an lvalue reference type (useful in template programming for "perfect forwarding") when an object initializer is a temporary and the compiler doesn't eliminate the copy/move entirely.
As shown above “move constructor” is faster than “copy constructor” as move constructor is using “char *” of object “mv1” to object “mv3”. Because of this performance of move constructor is better than copy constructor.
std::move () is a function used to convert an lvalue reference into the rvalue reference. Used to move the resources from a source object i.e. for efficient transfer of resources from one object to another. std::move () is defined in the <utility> header.
Calling a std::move () on a const object usually has no effect. It doesn’t make any sense to steal or move the resources of a const object. Copy semantics is used as a fallback for move semantics if and only if copy semantics is supported.
There is no move construction going on. std::move()
accepts a reference and returns a reference. std::move()
is basically just a cast.
Your guess 4. is the right one (assuming that you are actually calling a move constructor in the end).
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