Converting possible lvalue reference template to lvalue

Question

What I'm trying to achieve is something like this:

template<typename T> void foo(T t) { TWithoutReferenceType t1 = Magic(t); //some magic here }

TWithoutReference is the same type of T but without reference, so for example:

Magic(int i) -> int i Magic(int& i) -> int i Magic(int&& i) -> int i

I don't know if this is possible for rvalue references or what it could mean in practice though.

WHY I NEED IT:

template <typename ReturnType, typename... Args> function<ReturnType(Args...)> memoize(const function<ReturnType(Args...)>& func) { return ([=](Args... args) mutable { static map<tuple<Args...>, ReturnType> cache; tuple<Args...> t(args...); auto result = cache.insert(make_pair(t, ReturnType{})); ...

So for example if Args ... is vector<double> & and I change the order of the elements, the correspondent key in cache is changed also, and I don't want that this happens, so I don't want that a key change in the future as side effect.

Sorry if I posted the whole code, but I'm afraid that if I make it simpler I'll lose the point of the question.

Are you saying that you want tuple<Args...> to be a tuple of non-reference types — M.M
– M.M, Commented Apr 28, 2016 at 4:23
Exactly. And I think that this is a duplicate of this question — justHelloWorld
– justHelloWorld, Commented Apr 28, 2016 at 4:25
OK, let us know if that solution worked for you and I'll close the question — M.M
– M.M, Commented Apr 28, 2016 at 4:26

songyuanyao · Accepted Answer · 2016-04-28 08:14:43Z

What you're finding is std::remove_reference:

If the type T is a reference type, provides the member typedef type which is the type referred to by T. Otherwise type is T.

Possible implementation
template< class T > struct remove_reference {typedef T type;}; template< class T > struct remove_reference<T&> {typedef T type;}; template< class T > struct remove_reference<T&&> {typedef T type;}; 

You might implement Magic as

template <typename T> typename remove_reference<T>::type Magic(T t) { return t; }

So can you write the function that I'm look for using std::remove_reference please?

Jarod42 · Accepted Answer · 2016-04-28 08:33:21Z

2

You have several choices including:

template <typename T> typename std::remove_reference<T>::type Magic(const T& t) { return t; }

or

template <typename T> typename std::remove_reference<T>::type Magic(T&& t) { return t; }

edited Apr 28, 2016 at 8:33

answered Apr 28, 2016 at 8:06

Jarod42

228k15 gold badges214 silver badges351 bronze badges

6 Comments

M.M Over a year ago

you probably want typename std::remove_reference<T>::type as the return type in the second one.

songyuanyao Over a year ago

@M.M Isn't it the same thing?

Jarod42 Over a year ago

@songyuanyao: Not if you provide the template parameter, i.e: Magic<const int&>(my_int);, but that change nothing for Magic(my_int);

songyuanyao Over a year ago

@Jarod42 Sorry didn't get it. I meant typename std::remove_reference<T>::type and std::remove_reference_t<T> is the same thing, isn't it?

Jarod42 Over a year ago

There are the same but remove_reference_t is only available since c++14 (and OP's tag is c++11).

|

Collectives™ on Stack Overflow

Converting possible lvalue reference template to lvalue

2 Answers 2

1 Comment

6 Comments

Linked

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

1 Comment

6 Comments

Linked

Related