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".
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);