template type deduction of template parameter

Question

I know, that there exists the possibility for automatic type deduction of function templates, given a particular function parameter, but does there also exist such a method for non type template parameters?

Example:

#include <iostream>

template<typename T, T val>
void func_a(void) {
    std::cout << val << std::endl;
}

template<typename T>
void func_b(T val) {
    std::cout << val << std::endl;
}

int main(void) {
    func_a<uint32_t, 42u>();
    //func_a<42u>();    //This line doesn't work
    func_b(42u);
    return 0;
}

So I don't want to give each time the template argument type uint32_t every time, when I call func_a() . Does there such a method exist in C++17 or below?

I am using g++ v.7.3 and c++17.

Answer 1

In C++17, you can use auto :

template<auto val>
void func_a(void) {
    std::cout << val << std::endl;
}

int main(void) {
    func_a<42u>();
    return 0;
}

Answer 2

Given a +1 for the C++17 solution, a better-than-nothing C++11/C++14 solution can be the use of a macro to activate decltype() over the argument.

By example, with the macro

#define func_a_macro(val)  func_a<decltype(val), val>

or better, as suggested by liliscent, to avoid problems with references

#define func_a_macro(val) \
   func_a<std::remove_reference<decltype(val)>::type, val>

you can call

func_a_macro(42u)();

ps: I know... I know... macros are distilled evil... but sometime are useful.

Answer 3

C++14 solution with no macros:

template<int N> auto Int = std::integral_constant<int, N>{};

template<class T, T n>
constexpr auto foo(std::integral_constant<T, n> x)
{
    std::cout << x.value << std::endl;
} 

int main()
{
    foo(Int<6>);
}

c++11:

template<int N> using Int = std::integral_constant<int, N>;

template<class T, T n>
constexpr void foo(std::integral_constant<T, n> x)
{
    std::cout << x.value << std::endl;
} 

int main()
{
    foo(Int<6>());
}