Using enable_if to optionally add a struct member

c++, enable-if, metaprogramming

Solution

Well: use a base class.

struct Empty {};

struct SpecialTypeCnt { SpecialType member; };

template <typename A>
struct Something: if_< /* cond */ , SpecialTypeCnt, Empty>::type {
};

Where `if_` is defined as:

template <typename, typename, typename E> struct if_ { typedef E type; };

template <typename T, typename E>
struct if_<std::true_type, T, E> { typedef T type; };

(You can also specialize on a boolean)

Now of course, you need to express your condition properly.

Having said that, you should probably not use just a `struct`. Instead you should use a `class` which provides the operations that need be applied on `member`. Then you provide a `class Null` with a default behavior and a `class SomeType` with the behavior specific to `member`.

Otherwise you'll rewrite the condition everywhere you need to "perhaps" modify `member`, and it gets annoying real quick.

Problem

Given this template: ``` template <class A> struct Something { ... // members common to all template instantiations for all A types SpecialType member; // but not this - I want this to be conditional... } ``` ...I want to use "enable_if" to have the SpecialType member exist conditionally; that is, only when the template is instantiated with A=SpecialCase1 or SpecialCase2 types. In all other cases, I want the SpecialType member to be missing. In case you're wondering why, this is about optimization - i.e. not carrying useless payload in the struct. I am a newbie in template metaprogramming, but I understand I need "enable_if" and two "is_same" somehow - not sure exactly how, though... EDIT: Doing it with generic C++ (i.e. without Boost-specifics) would be a plus.

Original source