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

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.

Evg · Accepted Answer · 2019-11-22T09:25:02.573

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).

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

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