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++.

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.