How to pass a template function in a template argument list

Question

Suppose I have a template function:

template<typename T>
T produce_5_function() { return T(5); }

How can I pass this entire template to another template?

If produce_5_function was a functor, there would be no problem:

template<typename T>
struct produce_5_functor {
  T operator()() const { return T(5); }
};
template<template<typename T>class F>
struct client_template {
  int operator()() const { return F<int>()(); }
};
int five = client_template< produce_5_functor >()();

but I want to be able to do this with a raw function template:

template<??? F>
struct client_template {
  int operator()() const { return F<int>(); }
};
int five = client_template< produce_5_function >()();

I suspect the answer is "you cannot do this".

Jesse Good · Accepted Answer · 2013-03-27 04:57:11Z

25

I suspect the answer is "you cannot do this".

Yes, that is the case, you cannot pass a function template as a template argument. From 14.3.3:

A template-argument for a template template-parameter shall be the name of a class template or an alias template, expressed as id-expression.

The template function needs to be instantiated before you pass it to the other template. One possible solution is to pass a class type that holds a static produce_5_function like so:

template<typename T>
struct Workaround {
  static T produce_5_functor() { return T(5); }
};
template<template<typename>class F>
struct client_template {
  int operator()() const { return F<int>::produce_5_functor(); }
};
int five = client_template<Workaround>()();

Using alias templates, I could get a little closer:

template <typename T>
T produce_5_functor() { return T(5); }

template <typename R>
using prod_func = R();

template<template<typename>class F>
struct client_template {
  int operator()(F<int> f) const { return f(); }
};

int five = client_template<prod_func>()(produce_5_functor);

edited Mar 27, 2013 at 4:57

answered Mar 27, 2013 at 4:10

Jesse Good

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2 Comments

Olumide Over a year ago

Is there a fundamental reason why template template parameters cannot be function templates? Is this likely to be addressed in the future?

Ben Voigt Over a year ago

@Olumide: Probably because it's a lot of additional complexity and easy to work around (see mfontanini's answer). Same goes for partial specialization, which is used a whole lot more than template template arguments.

mfontanini · Accepted Answer · 2013-03-27 03:51:48Z

6

How about wrapping that function?

template<typename T>
struct produce_5_function_wrapper {
    T operator()() const { return produce_5_function<T>(); }
};

Then you can use the wrapper instead of the function:

int five = client_template< produce_5_function_wrapper >()();

Using the template function alone will not work, there's no such thing as "template template functions".

answered Mar 27, 2013 at 3:51

mfontanini

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