In C++17 suppose I have a function-like object passed as a parameter to some template:
template<typename F>
void g(F f) {
auto x = f(/*...*/);
}
There are lots of different types F could be, such as a function pointer, a std::function, a lambda expression, and in fact any class type that implements operator()
.
Is there any way to get the function-like objects arity and type of its parameters and the type of its return type?
I mean, ultimately F could be a class that overloads operator()
with multiple different member functions, each with different arities, parameter types and return types - so there isn't a fully-general answer (unless there is some way to iterate that overload set, which I don't think there is).
But for the typical case where a function call expression involving f results in a single overload, is there a solution?
(also if there is any progress in C++20, worth mentioning too)
C++17's adds deduction guides for std::function
, which we can use to do the deduce the function signature of non-overloaded function-like objects:
template <typename R, typename... Args>
constexpr auto do_something_with_the_signature(std::function<R(Args...)>) {
// Assuming that you only care about the return type.
// Otherwise, you may want some kind of wrapper to extract the signature to
// avoid runtime cost
}
...
using some_type_computation =
decltype(do_something_with_the_signature(std::function(f)));
If you only wanted the return type, you could just use:
using result_type = typename decltype(std::function(f))::result_type;
If you want to avoid std::function
altogether because of the compile-time costs, you can implement your own version of the deduction guides for your own type (possibly as general as a function_traits
type trait). A sketch of how you might implement the deduction guides yourself can be seen in my answer here: https://stackoverflow.com/a/66038056/1896169
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