Variadict template that calls variadic template received as parameter

Question

I have a variadic template function that I want to use to conditionally add arguments to another variadic template function. This is a minimal example, but I can't get it to compile:

// Copyright 2019 Google LLC.
// SPDX-License-Identifier: Apache-2.0

#include <utility>

template<typename... Args>
void f(Args&&... args) {
  // does something with args
}

template<typename T, typename... Args>
void g(T&& t, int i, Args&&... args) {
  if (i != 0) t(i, std::forward<Args>(args)...);
  else t(std::forward<Args>(args)...);
}

int main() {
  g(f, 0, 0);
  return 0;
}

The output from clang for the code above is:

main.cpp:15:7: error: no matching function for call
      to 'g'
      g(f, 0, 0);
      ^
main.cpp:9:10: note: candidate template ignored:
      couldn't infer template argument 'T'
    void g(T&& t, int i, Args&&... args) {
         ^
1 error generated.

With macros, it would work like this (compiles if I replace the g function above with this macro):

// Copyright 2019 Google LLC.
// SPDX-License-Identifier: Apache-2.0

#define g(t, i, args...) \
  if((i) != 0) (t)((i), args); \
  else (t)(args);

Is there a way I can make this work without macros?

user7860670 · Accepted Answer · 2019-09-15 18:15:13Z

5

Directly passing f won't work because function pointer must point at specific function template instance that must be instantiated at a point of g invocation. Using template template template parameter it is possible to create function signature builder that can be passed as a parameter without prior instantiation:

#include <utility>

template<typename... Args>
void f(Args... args) {
  // does something with args
}

template<typename... Args> struct
get_f
{
    static constexpr auto & get() { return f<Args...>; }
};

template<template<typename...> typename T, typename... Args>
void g(int i, Args&&... args) {
  if (i != 0) T<int, Args...>::get()(i, std::forward<Args>(args)...);
  else T<Args...>::get()(std::forward<Args>(args)...);
}

int main() {
  g<get_f>(0, 0);
  return 0;
}

answered Sep 15, 2019 at 18:15

user7860670

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5 Comments

RenatoUtsch Over a year ago

Thanks, that makes a lot of sense. One question though, why can't I specify f<Args...> like you specify T<Args...>? Why doesn't it work as a function pointer to be referenced?

user7860670 Over a year ago

@RenatoUtsch My answer starts from explanation why passing f isn't working.

RenatoUtsch Over a year ago

Yes. This probably comes from my lack of formal knowledge, but why f<Args...> is not a "specific function template instance that is instantiated at the point of g invocation, while get_f<Args...> is a "specific class template instance instantiated at the point of g invocation"?

user7860670 Over a year ago

@RenatoUtsch f<Args...> is specific function template that can be passed around as a reference to function or as a pointer to function. However you are not passing f<Args...> into g, you are passing just f which is only the name of function template which can not be passed anywhere. get_f is only the name of class template but it can be passed at template parameter.

RenatoUtsch Over a year ago

So function templates can't be passed as template arguments but class templates can? Do you know if there is any reason for this? It seems a bit arbitrary.

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