Why does boost::variant not provide operator !=

Question

Given two identical boost::variant instances a and b, the expression ( a == b ) is permitted.

However ( a != b ) seems to be undefined. Why is this?

Answer 1

I think it's just not added to the library. The Boost.Operators won't really help, because either variant would have been derived from boost::operator::equality_comparable. David Pierre is right to say you can use that, but your response is correct too, that the new operator!= won't be found by ADL, so you'll need a using operator.

I'd ask this on the boost-users mailing list.

Edit from @AFoglia's comment:

Seven months later, and I'm studying Boost.Variant, and I stumble over this better explanation of the omission lists.

http://boost.org/Archives/boost/2006/06/105895.php

operator== calls operator== for the actual class currently in the variant. Likewise calling operator!= should also call operator!= of the class. (Because, theoretically, a class can be defined so a!=b is not the same as !(a==b).) So that would add another requirement that the classes in the variant have an operator!=. (There is a debate over whether you can make this assumption in the mailing list thread.)

Answer 2

This is a link to the answer from the author himself when this question was formulated on boost mailing list

Summarizing it, in the author opinion, implementing comparison operators (!= and <) would add more requirements on the types used to create the variant type.

I don't agree with his point of view though, since != can be implemented in the same way as ==, without necessarily hiding the possible implementations of these operators for each of the types making up the variant

Answer 3

Because it doesn't need to.

Boost has an operators library which defines operator!= in term of operator==