How to deduce template by struct's member - c++

Question

Suppose I have some structs and each of them holds one enum as a member.
I want to call a method of a struct but depending on a struct's member, like in the code example:
#include <iostream>
#include <string>
#include <type_traits>
enum class Type{
lowercase = 0,
uppercase
};
struct Low{
static constexpr Type cp = Type::lowercase;
};
struct Up{
static constexpr Type cp = Type::uppercase;
};
template<typename Case, typename=void>
struct Printer
{
void print_g(const std::string& s){
std::cout << "just s: " << s << std::endl;
}
};
template<typename X>
struct Printer<X, std::enable_if_t<X::cp == Type::lowercase, void>>
{
void print_g(const std::string& s){
std::cout << "lowercase " << std::nouppercase << s << std::endl;
}
};
template<typename X>
struct Printer <X, std::enable_if_t<X::cp == Type::uppercase, void>>
{
void print_g(const std::string& s){
std::cout << "uppercase " << std::uppercase << s << std::endl;
}
};
int main()
{
Printer<Low> pl;
pl.print_g("hello1");
Printer<Up> p2;
p2.print_g("hello2");
}
But this solution doesn't look quite elegant to me.
Especially the part typname=void in the first template.
Only then code compiles. Why is that the case?
And is there any better (more elegant) solution for this template specialization?

Answer 1

In C++17, you can use if constexpr:
template <typename X>
struct Printer
{
void print_g(const std::string& s)
{
if constexpr(X::cp == Type::lowercase)
{
std::cout << "lowercase " << std::nouppercase << s << std::endl;
}
else if constexpr(X::cp == Type::uppercase)
{
std::cout << "uppercase " << std::uppercase << s << std::endl;
}
else
{
std::cout << "just s: " << s << std::endl;
}
}
};
If you do not have access to C++17, consider these options:
Use a regular if...else statement. There's no code that needs to be conditionally compiled in your example.
Implement static_if in C++14. Here's a talk I gave that explains how to do it: Implementing static control flow in C++14

Answer 2

You can fully specialize Printer for Low and Up.
template<class Case>
struct Printer
{
void print_g(const std::string& s) {
std::cout << "just s: " << s << std::endl;
}
};
template<>
struct Printer<Low>
{
void print_g(const std::string& s) {
std::cout << "lowercase " << std::nouppercase << s << std::endl;
}
};
template<>
struct Printer<Up>
{
void print_g(const std::string& s) {
std::cout << "uppercase " << std::uppercase << s << std::endl;
}
};
Notice that the enum does not come into play at all. If you need to specialize for the enum, you can do that too.
template<Type Case>
struct PrinterHelper
{
void print_g(const std::string& s) {
std::cout << "just s: " << s << std::endl;
}
};
template<>
struct PrinterHelper<Type::lowercase>
{
void print_g(const std::string& s) {
std::cout << "lowercase " << std::nouppercase << s << std::endl;
}
};
template<>
struct PrinterHelper<Type::uppercase>
{
void print_g(const std::string& s) {
std::cout << "uppercase " << std::uppercase << s << std::endl;
}
};
template<class Case>
using Printer = PrinterHelper<Case::cp>;

Answer 3

I would likely just go with:
enum struct Casing
{
Lower,
Upper
};
template<Casing>
struct printer;
template<>
struct printer<Casing::Lower>
{
...
};
template<>
struct printer<Casing::Upper>
{
...
};