Extract C++ template parameters

Question

Although I'm doubtful, I'm curious as to whether it's possible to extract primitive-type template parameters from an existing type, perhaps using RTTI.

For example:

typedef std::bitset<16> WordSet;

Would it be possible to extract the number 16 in the above code without hard-coding it elsewhere? Compiler specific implementations are welcome, though I'm particularly interested in g++.

Riendeau

Answer 1

It's not possible in general to pick arbitrary template parameters.

However, the usual way you do it is this:

template<int N>
struct foo {
    static const int value = N;
};

and for types

template<typename T>
struct foo {
    typedef T type;
};

You can access it then as foo<39>::value or foo<int>::type.

If you have a particular type, you can use partial template specialization:

template<typename>
struct steal_it;

template<std::size_t N>
struct steal_it< std::bitset<N> > {
    static const std::size_t value = N;
};

The same principle is possible for type parameters too, indeed. Now you can pass any bitset to it, like steal_it< std::bitset<16> >::value (note to use size_t, not int!). Because we have no variadic many template paramters yet, we have to limit ourself to a particular parameter count, and repeat the steal_it template specializations for count from 1 up to N. Another difficulty is to scan types that have mixed parameters (types and non-types parameters). This is probably nontrivial to solve.

If you have not the type, but only an object of it, you can use a trick, to still get a value at compile time:

template<typename T>
char (& getN(T const &) )[steal_it<T>::value];  

int main() {
    std::bitset<16> b;
    sizeof getN(b); // assuming you don't know the type, you can use the object
}

The trick is to make the function template auto-deduce the type, and then return a reference to a character array. The function doesn't need to be defined, the only thing needed is its type.

Faultfinder

Nice trick. One thing though, you start by saying how it's not possible, but then proceed to show that it is! :) – Cayuse

@Richard Corden: I think he was saying it's not possible to do using RTTI – Riendeau

Richart Corden. I meant it's generally not possible to pick arbitrary template parameters. consider this: template<template<class, int> class, class, size_t> class foo; that trick won't get their types :) – Faultfinder

Answer 2

You can easily do this in C++11 using argument deduction and unevaluated contexts (note that demo uses C++14's variable template feature for convenience).

#include <type_traits>
#include <iostream>

template<int>
struct foo {};

template<int arg_N>
struct val {
    static constexpr auto N = arg_N;
};

template<template <int> typename T, int N>
constexpr auto extract(const T<N>&) -> val<N>;

template<typename T>
constexpr auto extract_N = decltype(extract(std::declval<T>()))::N;


int main() {
    std::cout << extract_N<foo<5>>;
}

Live demo

Rameau

Answer 3

I like Marc Garcia's answer because it shows how to extract the template parameter in a generic way, but I believe his example can be simpler:

#include <type_traits>
#include <iostream>

template<int>
struct MyType {};

template<template <int> typename T, int N>
constexpr int extract(const T<N>&) { return N; }

int main() {
    constexpr MyType<5> myObj;
    std::cout << extract(myObj);
}

Live demo

Stefan

Answer 4

In the case of std::bitset, you can just use the size() member function:

size_t sz = oh_my_word.size();  // sz is now 16

In the general case, you can define a member function which returns the size in a similar manner:

template <int N>
class Foo
{
public:
  int size() const { return N; }
};

Emersen

I'm talking about extracting it from third-party classes with no size() method or anything similar -- extracting it just knowing the type itself. The reason being that I don't want to have to make a dummy instance just to get the size of the typedef'd bitset. – Riendeau

It's not a particularly bad penalty for bitset, but there could be other types where instantiating a dummy type is undesirable or impractical. – Riendeau

Answer 5

As stated by other answers, for std::bitset you can get the size by using the size() member function, which should be the right choice, better than any other trick.

There were several proposals for the generic case, almost similar to the one I suggest below, but still I think this one is simpler:

template <template<std::size_t> typename T, std::size_t K>
auto extractSize(const T<K>&) {
    return K;
}

int main() {
    std::bitset<6> f1;
    std::bitset<13> f2;
    std::cout << extractSize(f1) << std::endl;
    std::cout << extractSize(f2) << std::endl;
}

Metachromatism

Answer 6

For very old C++ and integer types, a class enum replaces the non-working static const declaration.

template<int N> struct foo {
  enum{value=N};
};

Selfstarter