What is the most efficient way to pass a non generic function? - c++

Question

I am learning functional programming in C++. My intention is to pass a non generic function as argument. I know about the template method, however I would like to restrict the function signature as part of the API design. I worked out 4 different methods example on cpp.sh:
// Example program
#include <iostream>
#include <string>
#include <functional>
typedef int(functor_type)(int);
int by_forwarding(functor_type &&x) {
return x(1);
}
int functor_by_value(functor_type x) {
return x(1);
}
int std_func_by_value(std::function<functor_type> x) {
return x(1);
}
int std_func_by_forwarding(std::function<functor_type> &&x) {
return x(1);
}
int main()
{
std::cout << functor_by_value([](int a){return a;}); // works
std::cout << std_func_by_value([](int a){return a;}); // works
std::cout << std_func_by_forwarding(std::move([](int a){return a;})); // works
//std::cout << by_forwarding([](int a){return a;}); // how to move lambda with forwarding ?
}
Is any of the above attempts correct? If not, how do i achieve my goal?

Answer 1

(based on clarification from comments)
Signature can be restricted by using std::is_invocable:
template<typename x_Action> auto
functor_by_value(x_Action && action)
{
static_assert(std::is_invocable_r_v<int, x_Action, int>);
return action(1);
}
online compiler

Answer 2

Other alternative:
template <typename Func>
auto functor_by_value(Func&& f)
-> decltype(std::forward<Func>(f)(1))
{
return std::forward<Func>(f)(1);
}

Answer 3

however I would like to restrict the function signature as part of the API design.
So restrict it:
#include <functional>
#include <type_traits>
#include <iostream>
/// #tparam F is a type which is callable, accepting an int and returning an int
template
<
class F,
std::enable_if_t
<
std::is_convertible_v<F, std::function<int(int)>>
>* = nullptr
>
int myfunc(F &&x) {
return x(1);
}
int main()
{
auto a = myfunc([](int x) { std::cout << x << std::endl; return 1; });
// does not compile
// auto b = myfunc([]() { std::cout << "foo" << std::endl; return 1; });
}

Answer 4

As usual, this depends on how good your compiler is today, and how good it will be in the future.
Currently, compilers are not very good at optimizing std::function. Surprisingly, std::function is a complicated object that sometimes has to allocate memory to maintain stateful lambda functions. It also complicates matters that std::function has to be able to refer to member function, regular functions, and lambdas, and do it in a transparent manner. This transparency has a hefty runtime cost.
So, if you want the fastest possible code, you should be careful with std::function. For that reason the first variant is the fastest (on today's compilers):
int functor_by_value(functor_type x) {
return x(1);
}
It simply passes a pointer to a function.
When stateful lambdas are involved you have only two options. Either pass the lambda as a template argument, or convert to std::function. Hence, if you want the fastest code possible with lambdas (in today's compilers), you'd pass the function as a templated argument.
Since a lambda function may have a big state, passing it around may copy the big state (when copy elision is not possible). GCC will construct the lambda directly on the parameter list (with no copy), but a nested function will invoke a copy constructor for the lambda. To avoid that, either pass it by const reference (in that case it can't be mutable), or by rvalue reference:
template<class Func>
void run2(const Func & f)
{
std::cout << "Running\n";
f();
}
template<class Func>
void run(const Func & f)
{
run2(f);
}
int main()
{
run([s=BigState()]() { std::cout << "apply\n"; });
return 0;
}
Or:
template<class Func>
void run2(Func && f)
{
f();
}
template<class Func>
void run(Func && f)
{
run2(std::forward<Func>(f));
}
int main()
{
run([s=BigState()]() { std::cout << "apply\n"; });
return 0;
}
Without using references, the BigState() will be copied when the lambda is copied.
UPDATE:
After reading the question again I see that it wants to restrict the signature
template<typename Func,
typename = std::enable_if_t<
std::is_convertible_v<decltype(Func(1)), int>>>
void run2(const Func & f)
{
std::cout << "Running\n";
f();
}
This will restrict it to any function that can accept int (possibly with an implicit cast), and returns an int or any type that is implicitly cast to int. However, if you want to accept only function-like objects that accept exactly int and return exactly int you can see if the lambda is convertible to std::function<int(int)>