Template parameter default to a later one

Question

This link doesn't answer my question so I'll ask it here:

Basically I want to write a template function

template <typename Out, typename In>
Out f(In x);

Here I always need to specify Out when calling f. I don't want to do it every time, so I basically want

template <typename Out = In, typename In>
Out f(In x);

Which means if I don't specify Out, it will default to In. However, this is not possible in C++11.

So my question is, is there any way to achieve the effect:

1. calling f(t) will instantiate f<T,T>(t) or more generally f<typename SomeThing<T>::type, T>
2. calling f<U>(t) will instantiate f<U, T>(t)

Answer 1

I have a PERFECT solution here! f<const int&> won't work because a function can't return a reference to a temporary, not related to the techniques used here.

[hidden]$ cat a.cpp
#include <iostream>
#include <type_traits>
#include <typeinfo>
using namespace std;

template <typename Out, typename In>
Out f_impl(In x) {
  cout << "Out=" << typeid(Out).name() << " " << "In=" << typeid(In).name() << endl;
  return Out();
}

template <typename T, typename... Args>
struct FirstOf {
  typedef T type;
};

template <typename T, typename U>
struct SecondOf {
  typedef U type;
};

template <typename... Args, typename In>
typename enable_if<sizeof...(Args) <= 1, typename FirstOf<Args..., In>::type>::type f(In x) {
  typedef typename FirstOf<Args..., In>::type Out;
  return f_impl<Out, In>(x);
}

template <typename... Args, typename In>
typename enable_if<sizeof...(Args) == 2, typename FirstOf<Args...>::type>::type f(In x) {
  typedef typename FirstOf<Args...>::type Out;
  typedef typename SecondOf<Args...>::type RealIn;
  return f_impl<Out, RealIn>(x);
}

int main() {
  f(1);
  f(1.0);
  f<double>(1);
  f<int>(1.0);
  f<int>(1);
  f<const int>(1);
  f<int, double>(1);
  f<int, int>(1);
  f<double, double>(1);
}
[hidden]$ g++ -std=c++11 a.cpp
[hidden]$ ./a.out
Out=i In=i
Out=d In=d
Out=d In=i
Out=i In=d
Out=i In=i
Out=i In=i
Out=i In=d
Out=i In=i
Out=d In=d

Answer 2

You probably never want to specify In but rather have it deduced, right?

In this case you need to overload the function:

template <typename Out, In>
Out f(In x);

template <typename T>
T f(T x);

Call it:

f(42);
f<float>(42);

… but unfortunately that’s ambiguous for f<int>(42). No matter, we can use SFINAE to disable one of the overloads appropriately:

template <
    typename Out,
    typename In,
    typename = typename std::enable_if<not std::is_same<Out, In>::value>::type
>
Out f(In x);

template <typename T>
T f(T x);

In order to avoid redundancy in the implementation, let both functions dispatch to a common implementation, f_impl.

Here’s a working example:

template <typename Out, typename In>
Out f_impl(In x) {
    std::cout << "f<" << typeid(Out).name() <<
                 ", " << typeid(In).name() <<
                 ">(" << x << ")\n";
    return x;
}

template <
    typename Out,
    typename In,
    typename = typename std::enable_if<not std::is_same<Out, In>::value>::type
>
Out f(In x) {
    std::cout << "f<Out, In>(x):\t ";
    return f_impl<Out, In>(x);
}

template <typename T>
T f(T x) {
    std::cout << "f<T>(x):\t ";
    return f_impl<T, T>(x);
}


int main() {
    f(42);
    f<float>(42);
    f<int>(42);
}

Answer 3

You may not need it here, but here is a classical technique:

struct Default
{
  template <typename Argument, typename Value>
    struct Get {
      typedef Argument type;
    };

  template <typename Value>
    struct Get <Default, Value> {
      typedef Value type;
    };
};

template <typename Out = Default, typename In>
typename Default::Get<Out, In>::type f(In x);