decltype of function parameter

Question

Is it possible to deduce the type of a function parameter? For example, if I have:

void foo(int a);

I would like to deduce the type int as the type of foo's first parameter. A possible use could be:

foo( static_cast< decltype(/* ??? foo's first param ??? */) >(value) );

In this related question, the answers exploit having a member with the same type for deduction, so it does not directly deduce the function parameter type.

Answer 1

Is it possible to deduce the type of a function parameter?

Sure.

With a type traits, by example (argType)

template <typename>
struct argType;

template <typename R, typename A>
struct argType<R(A)>
 { using type = A; };


void foo(int a)
 { }

int main()
 {
   long value = 1L;

   foo( static_cast<typename argType<decltype(foo)>::type>(value) );
 }

If you're interrested in a little more generic solution, the following example show how create and use a type traits to detect the return type or the n-th argument type

#include <string>

template <std::size_t N, typename T0, typename ... Ts>
struct typeN
 { using type = typename typeN<N-1U, Ts...>::type; };

template <typename T0, typename ... Ts>
struct typeN<0U, T0, Ts...>
 { using type = T0; };

template <std::size_t, typename>
struct argN;

template <std::size_t N, typename R, typename ... As>
struct argN<N, R(As...)>
 { using type = typename typeN<N, As...>::type; };

template <typename>
struct returnType;

template <typename R, typename ... As>
struct returnType<R(As...)>
 { using type = R; };

long bar (int a, std::string const &)
 { return a; }

int main()
 {
   long         valI = 1L;
   char const * valS = "abc";

   bar( static_cast<typename argN<0U, decltype(bar)>::type>(valI),
        static_cast<typename argN<1U, decltype(bar)>::type>(valS) );

   static_assert(
      std::is_same<long,
                   typename returnType<decltype(bar)>::type>::value, "!");
 }