why std::move uses std::remove_reference? [duplicate]

Question

According to http://en.cppreference.com/w/cpp/utility/move

std::move is declared as follows:

template <typename T> std::remove_reference<T>::type&& move(T&& t);

As far as my understanding goes, when code is templated, the deduction of T in typename T looses information about reference, so following:

template <typename T> void someFunction(T&& value);

when used like:

int five=5; someFunction(five);

then

value is of type int&
T is int

or

const float value = 5.25; someFunction(value);

then

value is of type const float&
T is const float.

If this is so, then there is no point in move declaration to declare returned type as: std::remove_reference<T>::type&&, because T is already not a reference.

Moreover, if std::move takes as argument a reference (l-value reference in practice), then returning static_cast<T&&>(t) in std::move in fact due to reference collapsing will return l-value reference or r-value reference, so it will behave more like std::forward not move. So what is a trick, that makes it work properly, that I do not understand?

T must be int& if value is to be int&. Reference collapsing. And you don't want that to happen for the return type, you want it to be r-value reference at all times. — LogicStuff
– LogicStuff, Commented Feb 8, 2016 at 9:51
I once made this very same mistake. How embarrassing! :-) open-std.org/jtc1/sc22/wg21/docs/papers/2002/… — Howard Hinnant
– Howard Hinnant, Commented Feb 8, 2016 at 15:18

TartanLlama · Accepted Answer · 2016-02-08 09:51:22Z

8

Your examples are incorrect:

int five=5; someFunction(five);

In this case, T is deduced as int&, not int. The same goes for the second example; T is deduced as const int&.

As such, returning just T&& would mean T&& &, which is T& due to reference collapsing rules.

This is why the std::remove_reference is required, to ensure that there are no references on the type to prevent reference collapsing from taking place.

answered Feb 8, 2016 at 9:51

TartanLlama

66.2k13 gold badges167 silver badges203 bronze badges

Sign up to request clarification or add additional context in comments.

6 Comments

DawidPi Over a year ago

what would be T then, if I had a function following: template <typename T> void function(T& a, T&& b); ?

TartanLlama Over a year ago

@DawidPi and pass what to it?

DawidPi Over a year ago

yeah right sorry, int a, b, function(a,std::move(b));

TartanLlama Over a year ago

T is int in that case.

TartanLlama Over a year ago

@DawidPi Template deduction rules are pretty complicated. Essentially, references are usually not deduced for T, but there's a special case for T&& parameters which states that A& is used instead of A if the argument is an lvalue (where A is the type of the argument).

|

Collectives™ on Stack Overflow

why std::move uses std::remove_reference? [duplicate]

1 Answer 1

6 Comments

Linked

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

6 Comments

Linked

Related