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c++templatesvariadic-templates
Alexander
Alexander

Reputation: 420

Variadic template, function as argument

I would like to use a function as argument in a variadic template, why does the following not work? How do I make it work?

template<typename F, typename... Args>
F test(F f, const Args&&... args) {
return f(std::forward<Args>(args)...);
}

int simple(int i) { 
    return i; 
}


int main()
{
    std::cout << test(simple, 2); // error, 'std::forward': none of the 2 overloads could convert all the argument types
}

Upvotes: 0

Views: 1099

Answers (2)

Ivaylo Valchev
Ivaylo Valchev

Reputation: 10415

The first problem is the return type. Your test function returns F which is a function pointer. Instead change it to auto to automatically deduce the return type.

The second issue is that std::forward requires a non-const reference.

You might use trailing return type:

template<typename F, typename... Args>
auto test(F f, Args&&... args) -> decltype(f(std::forward<Args>(args)...)) {
    return f(std::forward<Args>(args)...);
}

But decltype(auto) (C++14 required) is a simpler solution:

template<typename F, typename... Args>
decltype(auto) test(F f, Args&&... args) {
    return f(std::forward<Args>(args)...);
}

Upvotes: 0

Edgar Rokjān
Edgar Rokjān

Reputation: 17483

There are a couple of problems with your code.

First of all, you should use forwarding references, so you need to change const Args&&... to Args&&....

Then, test does not have to return F. So it is reasonable to use decltype(auto) here.

In addition to that, it makes sense to forward f too.

The fixed version might look like this:

template<typename F, typename... Args>
decltype(auto) test(F&& f, Args&&... args) {
    return std::forward<F>(f)(std::forward<Args>(args)...);
}

WANDBOX EXAMPLE

Upvotes: 3

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