Feb. 2, 2012
4:17 p.m.
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 <watchlala@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