[boost] On bind::operator==, again

What do you think this expression should do: bind(f, _1) == bind(g, _1) Currently, boost::bind returns f == g when f and g are of the same type, and fails at compile type otherwise (this is the equality comparison feature Doug requested). lambda::bind returns lambda(x) { return f(x) == g(x) }. It is clear that a future boost::bind that has limited lambda support would be even more problematic (as would be a future equality comparable boost.lambda). In that case, the expression will always compile, but do something different depending on whether f and g are of the same type. -- Peter Dimov http://www.pdimov.com