Because to calculate the result, the block must always be evaluated, so it's a bit redundant.
-- Cheers, Peter. On 2 feb 2012, at 17:08, Robert Sirois <watchl...@hotmail.com> wrote: > Ok, this may be from a simple lack of understanding, but I just have to know > ;) > > Why can I call #or: (Boolean and friends) with a block, but #xor: only takes > a Boolean? I would think the paradigm would remain the same and be able to > compose operators like #or:, etc.? > > What am I missing? Thanks! > > RS
