Function calling any specified function

Question

How to write a function calling any specified function (or function object) with specified arguments?

Here's what I tried:

#include <iostream>
#include <functional>

using namespace std;

template <typename RetType, typename... ArgTypes>
RetType q(function<RetType(ArgTypes...)> f, ArgTypes... args)
{
    return f(args...);
}

int h(int a, int b, int c) { return a + b + c; }

int main()
{
    auto r = q(h, 1, 2, 3);

    cout << "called, result = " << r;

    return 0;
}

The compiler says, that template argument deduction/substitution failed due to mismatched types 'std::function<_Res(_ArgTypes ...)>' and 'int (*)(int, int, int)'.

I'm not sure why template arguments can't be deduced in my code.

You can try to use `make_function` on `h`: http://stackoverflow.com/a/8670836/2639398 — Curve25519, Jul 15 '15 at 20:16

score 3 · Accepted Answer · answered Jul 15 '15 at 20:16

3

Since it's a template anyway, you don't need std::function at all. Just do this:

template <class F, class... Arg>
auto q(F f, Arg... arg) -> decltype(f(arg...))
{
  return f(arg...);
}

Even better, use perfect forwarding:

template <class F, class... Arg>
auto q(F f, Arg&&... arg) -> decltype(f(std::forward<Arg>(arg)...))
{
  return f(std::forward<Arg>(arg)...);
}

answered Jul 15 '15 at 20:16

Angew is no longer proud of SO

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1

Since C++14, specifying trailing return type is unnecessary, thanks to return type deduction, so `auto q(F f, Arg&&... arg)` is enough. – user3900460 Jul 16 '15 at 09:39

score 0 · Answer 2 · answered Jul 15 '15 at 20:26

You're probably taking a difficult path for a simple problem... You can just use closures taking no parameters for that...

#include <iostream>
#include <functional>

using namespace std;

int h(int a, int b, int c) { return a + b + c; }

int main()
{
    auto clsr = [](){ return h(1, 2, 3); };

    auto r = clsr();
    cout << "called, result = " << r;
    return 0;
}

with the advantage that the IDE will even suggest arguments for h when writing/editing this kind of code.

Function calling any specified function

2 Answers2