non-capturing lambda and function pointer as parameter in overloaded function ambiguity - c++

Question

#include <functional>
#include <iostream>
template<typename T>
void test( std::function< void ( const T& ) > f )
{
T val {};
f( val );
std::cout << "std::function" << std::endl;
}
template<typename T>
void test( void(*f) ( const T& ) )
{
T val {};
f( val );
std::cout << "function pointer" << std::endl;
}
int main()
{
auto capturing_var { 0 };
// Works because implicit conversion to function pointer isn't applied when lambda is capturing
test< int >( [ capturing_var ]( const int& x ) { } );
// Doesn't work because implicitly convertible to function pointer and makes ambiguity
// I want this line to work somehow, how can i make it worked with same client code ? Is it possible ?
test< int >( []( const int& x ) { } );
// This line is finer if it works, but in this case compiler cannot deduce T and also ambiguous if it could
test( []( const int& x ) { } );
// Works because of unary + operator so conversion to function ptr and i dont need to specify T
test( +[]( const int& x ) { } );
return 0;
}
Actually I just want this code to work without changing anything in main(), but I'm not sure whether it is possible.
I can omit overloaded function which takes function pointer, then it would work but in that case I need to specify what T is.
Note: I need T to be deduced.

Answer 1

You can use technique described in this question with some small postprocessing.
Minimal example:
template<typename Ret, typename Arg>
Arg argument_type(Ret(*)(Arg));
template<typename Ret, typename Fn, typename Arg>
Arg argument_type(Ret(Fn::*)(Arg) const);
template<typename Fn>
auto argument_type(Fn) -> decltype(argument_type(&Fn::operator()));
template<typename Arg, typename Fn>
struct argument {
static_assert(std::is_invocable_v<Fn, Arg>);
using type = Arg;
};
template<typename Fn>
struct argument<void, Fn>{
using type = decltype(argument_type(std::declval<Fn>()));
};
template<typename T = void, typename Fn>
void test(Fn fn) {
using Arg = std::decay_t<typename argument<T, Fn>::type>;
std::cout << "Arg = " << boost::typeindex::type_id_with_cvr<Arg>().pretty_name()
<< std::endl;
}
int main() {
int capturing_var;
test<int>([capturing_var](const int& x) {}); // Arg = int
test<int>([](const int& x) {}); // Arg = int
test([](const int& x) {}); // Arg = int
test(+[](const int& x) {}); // Arg = int
test<int>([](auto x) {}); // Arg = int
}
If the argument type cannot be deduced, e.g., for a variadic lambda, it has to be provided (last example).