CODEWITHSUNDEEP
c++forwardingperfect
greg_p
greg_p

Reputation: 324

calling perfectly forwarded lambda

I have the following class which stores a perfectly forwarded reference to a function:

template <class F> 
class Visitor
{
public:
    Visitor(F &&f) : _f(std::forward<F>(f)) 
    {
    }

    xt::LOOP_CONTROL OnElem(xvi::Elem &elem)
    {
        return _f(elem);

        // or should it be: 
        // return std::forward<F>(_f)(elem);
    }

private:
    F &&_f;
};

Wondering if it makes a difference to call _f(elem) or std::forward<F>(_f)(elem);?

Upvotes: 0

Views: 178

Answers (1)

Jarod42
Jarod42

Reputation: 218238

Lambda cannot have ref qualified operator () currently, so for them, both forms are equivalent.

For custom functor, it might differ with std::forward<F>(_f)(elem);:

struct MyFunctor
{
    xt::LOOP_CONTROL operator()(xvi::Elem &elem) &  { std::cout << "lvalue"; return {};}
    xt::LOOP_CONTROL operator()(xvi::Elem &elem) && { std::cout << "rvalue"; return {};}
};

xvi::Elem elem;
Visitor<MyFunctor>(MyFunctor{}).OnElem(elem); // rvalue
MyFunctor func;
Visitor<MyFunctor&>(func).OnElem(elem); // lvalue

Upvotes: 2

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