C++ Understanding the exact semantics of templates, esp. with functional parameters - c++

Question

I'm comfortable using templates when they simply parameterize over a type. However, I'm now beginning to use them in more complex applications (specifically, parameterizing over a function, as in C++ Template: typename and function to map to int ), and have been reduced essentially to trial and error (note in the above referenced question I can't get my code to compile).
What are the exact semantics of the template keyword (or the typename keyword when used outside the definition of a template), especially when you want a function as a parameter?
How do I define a template that takes a function as a parameter?
Once defined, how do I instantiate and use that template?
I'll accept the first complete answer that includes both a compilable example as well as clear, full explanation, of what's going on behind the magic?
CLARIFICATION: I understand how to use typename. My question is: What are the exact semantics of template definition, especially when applied to invocable parameters? and How do you define and instantiate a class which takes a function (or functor or something invokable) as a template parameter?

Answer 1

What are the exact semantics of the template and typename keywords, especially when you want a function as a parameter?
template introduces a template declaration (and also declarations of explicit instantiations and specialisations, but that's beyond the scope of this question). It's followed by the list of template parameters, in angle brackets <>.
Within that list, typename or class introduces a type parameter; one of three kinds of parameter, the others being non-type (value) parameters, and template parameters.
How do I define a template that takes a function as a parameter?
The simplest way is with a type parameter, which can be any type that can be called like a function; for example
template <typename F, typename... Args>
void call(F && f, Args &&... args) {f(std::forward<Args>(args)...);}
Once defined, how do I instantiate and use that template?
If it's a function template, simply call the function with appropriately typed arguments. The template parameters will be deduced from them:
void f1(int a, double b); // function
struct fn {
void operator()(std::string);
};
call(f1, 42, 1.5);
call(fn{}, "Hello");
call([](std::string s){std::cout << s}, "Lambda\n");
For class templates, you'd have to specify the function type, which can be a bit messy:
template <typename F>
struct thing {
F f;
};
thing<void(*)(int,double)> thing1 {f1};
thing<fn> thing2 {fn{}};
// Lambdas are problematic since the type has no name
//thing<???> thing3 {[]{std::cout << "Lambda\n";}};
The messiness could be avoided by using a factory function template to deduce the function type and instantiate the class template:
template <typename F>
thing<F> make_thing(F && f) {
return thing<F>(std::forward<F>(f));
}
auto thing1 = make_thing(f1);
auto thing2 = make_thing(fn{});
auto thing3 = make_thing([]{std::cout << "Lambda\n";});

Answer 2

Below some example (commented) code to show how functions can be used in a template. You can replace typename Parameter with typename... Parameter and p to p... if you want the first template to work for any number of fucntion arguments, which uses variadic templates.
#include <functional> // for std::function
#include <iostream>
// example functions
void func(int i)
{
std::cout << "Integer: " << i << ".\n";
}
int gunc(int i)
{
int result = i+1;
std::cout << "+Integer:" << result << ".\n";
return result;
}
// general non-restrictive template
template<typename Function, typename Parameter>
void call_twice(Function f, Parameter p)
{
f(p);
f(p);
}
// Restrict signature to void(int), but the return value can be anything: it will be ignored
void call_thrice(std::function<void(int)> f, int p)
{
f(p);
f(p);
f(p);
}
// Restrict signature to int(int), to exclude func
void call_four_times(std::function<int(int)> f, int p)
{
f(p);
f(p);
f(p);
f(p);
}
int main()
{
call_twice(func, 1); // instantiates void call_twice(void (*)(int), int)
call_twice(gunc, 1); // instantiated void call_twice(int (*)(int), int)
// Note I do not write the explicit types of func and gunc in the above comments, they're not important!
call_thrice(func, 10); // converts func to a std::function<void(int)>
call_thrice(gunc, 10); // return value can be ignored, so int(int) is convertible to void(int)
//call_four_times(func, 100); // will fail to compile: cannot convert a function of signature void(int) to one with signature int(int)
call_four_times(gunc, 100); // converts gunc to std::function<int(int)>
}
Live demo here.
The basic form of a template declaration starts anything of the form
template<typename T>
template<class T> // same as above
template<int N> // explicitly typed template parameter, need not be an int
template<typename... variadic_template_args>
And of course the forms with combinations of the above, including nested template types and all. Just remember the variadic template parameter must come last.
Templates are what the name implies: a blueprint to create a class with certain functionality expressed within the template definition.
Instantiation occurs when you actually call a template function (from either another instantiated template function or a non-template function), or for class templates, when you declare an object of the type, e.g. for std::vector:
std::vector<double> v; // instantiate the class template std::vector for type double
This is no different to templates that take functions as arguments. Note that functions decay to function pointers when passed to another function.
For example, a class template with a function as template parameter:
#include <iostream>
#include <type_traits> // for std::result_of
template<typename Function, Function f>
struct A
{
using function_type = Function;
template<typename... ArgTypes>
typename std::result_of<Function(ArgTypes...)>::type call(ArgTypes... args)
{
return f(args...);
}
};
void func(int i) { std::cout << "Integer: " << i << ".\n"; }
int main()
{
A<decltype(&func), func> a;
a.call(42);
}
Live demo here. Note the extra & in the decltype, because a function only decays to a function pointer when it is passed to a function, not when it is used as a template parameter. Note the "extra" typename before the std::result_of: this is required because its type typedef is a dependent name. These and other times you need to use the template and typename keywords, those are covered in Where and why do I have to put the “template” and “typename” keywords?

Answer 3

It's easier to use lambdas
template <typename F>
void herp(F function)
{
function("Hello from herp");
}
int main()
{
const auto deDerp = [] (const std::string& msg)
{
std::cout << msg << "\n";
};
herp(deDerp);
}