Use a lambda as a parameter for a C++ function

Question

I would like to use a lambda as a parameter for a C++ function, but I don\'t know which type to specify in the function declaration. What I would like to do is this:

Answer 1

You have two choices, basically.

Make it a template:

template<typename T>
void myFunction(T&& lambda){
}

or, if you do not want (or can't) do that, you can use type-erased std::function:

void myFunction(std::function<void()> const& lambda){
}

---

Conversely, your attempt with auto would've been correct under the concepts TS as currently implemented in gcc, where it'd be an abbreviated template.

// hypothetical C++2x code
void myFunction(auto&& lambda){
}

or with a concept:

// hypothetical C++2x code
void myFunction(Callable&& lambda){
}

Answer 2

You have 2 ways: make your function template:

template <typename F>
void myFunction(F&& lambda){
    //some things
}

or erase type with std::function

void myFunction(const std::function<void()/* type of your lamdba::operator()*/>& f){
    //some things
}

Answer 3

If this is an inline function, prefer a template, as in

template<typename Func>
void myFunction(Func const&lambda)
{
    //some things
}

because it binds to anything that makes sense (and will cause compiler error for anything else), including lambdas, instances of named classes, and std::function<> objects.

On the other hand, if this function is not inline, i.e. implemented in some compilation unit, you cannot use a generic template but must use a specified type, which is best taken a std::function<> object and passed via reference.