C++ template black magic

Question

This needs only work in g++.

I want a function

template<typename T> std::string magic();

such that:

Class Foo{}; magic<Foo>(); // returns "Foo";
Class Bar{}; magic<Bar>(); // returns "Bar";

I don't want this to be done via specialization (i.e. having to define magic for each type. I'm hoping to pull some macro/template black magic here. Anyone know how?)

Thanks!

Answer 1

The stringizing operator on macros may be what you're looking for:

#define MAGICCLASSNAME(str) std::string magic(#str)
class Foo{}; MAGICCLASSNAME(foo)

Answer 2

Try typeid(Foo).name() for a start. Parse as you see fit; will be implementation-dependent (but simply getting a string back is portable).

Answer 3

I have come up with the below:

#include <iostream>
#include <string>
#include <typeinfo>

using namespace std;
class Foo{}; 
class Bar{};

template<typename T> 
inline std::string magic(const T& obj)
{
 return typeid(obj).name();
}

int main()
{
 Foo a;
 cout << magic<Foo>(a); // returns "Foo";
}

I tested this with g++ and works well.

Also I got it from this SO answer.

Answer 4

To convert a type (or other identifer) into a string you need a macro, but a macro can not check if it's parameter is a valid type. To add type checking a template function can be added to the macro:

template<typename T>
std::string magic_impl(const char *name) { return name; }

#define more_magic(a) magic_impl<a>(#a)
#define magic(a) more_magic(a)

Here magic(int) gives the string "int" while magic(Foo) gives a "‘Foo’ was not declared" error if there is no such class.

Answer 5

In C++ probably you are looking template black magic. If so, Modern C++ Design will be helpful.