Question

Can I have a class that either has or not has member based on template? Imaginary code:

template <typename HasBazMember=true>
class Foo {
  int bar;

  ConditionallyHaveMember<int, HasBazMember> baz;
};

So in above I want Foo to have member “baz” and Foo not.

Answer 1

Yes, you can, but you have to specialize the entire class. Example:

template< bool HasBazMember = true >
class Foo {
  int bar;
  int baz;
};

template<>
class Foo< false > {
  int bar;
};

If you can separate the logic, you can put those members in a base class so that you don’t need to duplicate the code for the entire class. For instance,

template< bool HasBazMember >
class FooBase
{
protected:
   int baz;
};

template<>
class FooBase< false >
{
    // empty class
    // the Empty Base Optimization will make this take no space when used as a base class
};

template< bool HasBazMember = true >
class Foo : FooBase< HasBazMember >
{
    int bar;
};

Or using Boost.CompressedPair:

struct empty {};

template< bool HasBazMember = true >
class Foo
{
    boost::compressed_pair<
        int
      , typename std::conditional< HasBazMember, int, empty >::type
    > bar_and_maybe_baz_too;
};