Enable class constructor in some enumerated template cases

Question

For performance reasons, I'm using a templated class with an enum instead of relay on inheritance (It is not an option).

At this point I have something like:

typedef enum { A, B, C, D } QueueType;

template <QueueType T> class Queue {
    Queue(int a){...} // only usable when T = A
    Queue(unsigned a, unsigned b){...} // only usable when T = B || T = C
    Queue(somestruct z){...} // only usable when T = B || T = C
    //other constructors
}

Now I'm using an annoying amount of ifs/switches over T and rising exceptions if an incompatible constructor is called for a defined T.

What I want is to use std::enable_if or equivalent to prevent throwing exceptions on constructor and detect on compilation time such kind of errors.

I've tried many stack-overflows and foreign sites std::enable_if examples, but I can barely understand what I'm really doing and I always end on a compilation error.

Thanks in advance and sorry for asking for a probably trivially answered question. I'm noob with templates.

Environment: Linux GCC 8 and c++14 Restrictions: Maximal performance with no virtual methods.

Answer 1

What I want is to use std::enable_if or equivalent to prevent throwing exceptions on constructor and detect on compilation time such kind of errors.

I've tried many stack-overflows and foreign sites std::enable_if examples, but I can barely understand what I'm really doing and I always end on a compilation error.

The problem with std::enable_if (and SFINAE, more in general) is that it works only checking template parameters. So can enable/disable a full class, with a test over a template parameter of the class, but can't enable/disable a single method, with a test over a template parameter of the class.

If you want SFINAE enable/disable a method (like your constructors) you have to made it a template method and test a template parameter of the method itself.

So you can't write something as

template <typename = std::enable_if_t<T == A>>
Queue (int)
 { } // only usable when T = A

because T is a template parameter of the class, not of the constructor.

But there is a trick: you can use default values/types for template parameters; so the following code works

template <QueueType U = T, typename = std::enable_if_t<U == A>>
Queue (int)
 { } // only usable when T = A 

because is checked the value U that is a template parameter of the constructor.

To enable the second constructor only when T is B or C, you can write

template <QueueType U = T, typename = std::enable_if_t<(U == B) || (U == C)>> 
Queue (unsigned, unsigned)
 { } // only usable when T = B || T = C

The following is a full compiling example

#include <type_traits>

typedef enum { A, B, C, D } QueueType;

template <QueueType T>
struct Queue
 {
   template <QueueType U = T, typename = std::enable_if_t<U == A>>
   Queue (int)
    { } // only usable when T = A

   template <QueueType U = T, typename = std::enable_if_t<(U == B) || (U == C)>>
   Queue (unsigned, unsigned)
    { } // only usable when T = B || T = C
 };

int main()
 {
   Queue<A>  qa0{1};         // compile
   //Queue<A>  qa1{1u, 2u};  // compilation error

   // Queue<B>  qb0{1};      // compilation error
   Queue<B>  qb1{1u, 2u};    // compile

   // Queue<C>  qc0{1};      // compilation error
   Queue<C>  qc1{1u, 2u};    // compile

   // Queue<D>  qd0{1};      // compilation error
   // Queue<D>  qd1{1u, 2u}; // compilation error
 }