CODEWITHSUNDEEP
c++templatesenumstypedef
SLC
SLC

Reputation: 2207

error: T does not name a type - For specialisation using strongly typed enums

I was trying to avoid duplicating code for by having a base class with mostly static functions and members. And then I'd derive from the base class in some other classes while using the code from the base class.

#include <unordered_set>

enum struct Type
{
    A = 0, B, C
};

template <Type T> struct Impl;

template <Type T> struct Base
{
    typedef T Tp; // error: 'T' does not name a type

    typedef Impl<T> Imp;

    static std::unordered_set<Imp*> _Inst;

    static void grab(Imp * ptr) { _Inst.insert(ptr); }
    static void drop(Imp * ptr) { _Inst.erase(ptr); };
    static void swap(Imp * ptr, Imp * old) { drop(old); grab(ptr); }

    // ...
};

template <> struct Impl<Type::A> : public Base<Type::A>
{
    // ...
};

template <> struct Impl<Type::B> : public Base<Type::B>
{
    // ...
};

template <> struct Impl<Type::C> : public Base<Type::C>
{
    // ...
};

int main(int argc, char** argv)
{
    return 0;
}

The actual implementation is a lot more different than that example and requires me to have an approach similar to that.

Upvotes: 0

Views: 484

Answers (1)

Pradhan
Pradhan

Reputation: 16737

T isn't a type and you can only typedef types. Create a member variable. For example,

constexpr Type Tp = T;

Upvotes: 1

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