c++ - Extracting typename from under typename - Stack Overflow

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I have the following method in a class A in C++ 11 where I know that TMethod gets converted to a typename B<C>. How can I extract the C typename so that I can access further members from it like in the example below?

template<typename TMethod, typename... Args>
decltype(auto) calledMethod(const Member<TMethod>& method)
{ 
    // somehow extract C from under TMethod
    C::Output dummy;
    return dummy;
}

I have the following method in a class A in C++ 11 where I know that TMethod gets converted to a typename B<C>. How can I extract the C typename so that I can access further members from it like in the example below?

template<typename TMethod, typename... Args>
decltype(auto) calledMethod(const Member<TMethod>& method)
{ 
    // somehow extract C from under TMethod
    C::Output dummy;
    return dummy;
}

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Share Improve this question Follow asked 9 hours ago JustplayitJustplayit 71711 gold badge77 silver badges2323 bronze badges 3

• Is B always the same type? – Drew Dormann Commented 9 hours ago
• 2 Are you perhaps looking for something like template<template <typename> class B, typename C, typename... Args> decltype(auto) calledMethod(const Member<B<C>>& method) ? – Igor Tandetnik Commented 9 hours ago
• I would say minimal reproducible example is needed. Do not describe code, just provide it in minimal complete verifiable version. Something we could just copy paste and try compile/run. – Marek R Commented 9 hours ago

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1 Answer 1

Sorted by: Reset to default 0

I like using what I call a meta function for this. Having

template<typename T, typename C>
auto get_template_type(T<C>) -> C;

we can write an alias to get C like

template<typename T>
using template_type_t = decltype(get_template_type(decval<T>()));

and then in your function you you can get C like

template<typename TMethod, typename... Args>
decltype(auto) calledMethod(const Member<TMethod>& method)
{ 
    using C = template_type_t<TMethod>;
    C::Output dummy;
    return dummy;
}

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