Should templated functions take lambda arguments b

Question

GCC 4.7 in C++11 mode is letting me define a function taking a lambda two different ways:

// by value
template<class FunctorT>
void foo(FunctorT f) { /* stuff */ }

And:

// by r-value reference
template<class FunctorT>
void foo(FunctorT&& f) { /* stuff */ }

But not:

// by reference
template<class FunctorT>
void foo(FunctorT& f) { /* stuff */ }

I know that I can un-template the functions and just take std::functions instead, but foo is small and inline and I'd like to give the compiler the best opportunity to inline the calls to f it makes inside. Out of the first two, which is preferable for performance if I specifically know I'm passing lambdas, and why isn't it allowed to pass lambdas to the last one?

Answer 1

FunctorT&& is a universal reference and can match anything, not only rvalues. It's the preferred way to pass things in C++11 templates, unless you absolutely need copies, since it allows you to employ perfect forwarding. Access the value through std::forward<FunctorT>(f), which will make f an rvalue again if it was before, or else will leave it as an lvalue. Read more here about the forwarding problem and std::forward and read here for a step-by-step guide on how std::forward really works. This is also an interesting read.

FunctorT& is just a simple lvalue reference, and you can't bind temporaries (the result of a lambda expression) to that.

Answer 2

This is a good question -- the first part: pass-by-value or use forwarding. I think the second part (having FunctorT& as an argument) has been reasonably answered.

My advice is this: use forwarding only when the function object is known, in advance, to modify values in its closure (or capture list). Best example: std::shuffle. It takes a Uniform Random Number Generator (a function object), and each call to the generator modifies its state. The function object is forwarded into the algorithm.

In every other case, you should prefer to pass by value. This does not prevent you from capturing locals by reference and modifying them within your lambda function. That will work just like you think it should. There should be no overhead for copying, as Dietmar says. Inlining will also apply and references may be optimized out.

Answer 3

When you create a lambda function you get a temporary object. You cannot bind a temporary to a non-const l-value references. Actually, you cannot directly create an l-value referencing a lambda function.

When you declare you function template using T&& the argument type for the function will be T const& if you pass a const object to the function, T& if you pass a non-const l-value object to it, and T if you pass it a temporary. That is, when passing a temporary the function declaration will take an r-value reference which can be passed without moving an object. When passing the argument explicitly by value, a temporary object is conceptually copied or moved although this copy or move is typically elided. If you only pass temporary objects to your functions, the first two declarations would do the same thing, although the first declaration could introduce a move or copy.