From: Brian Adkins <lojicdotcom@...>
Date: 2008-04-09T00:45:06+09:00
Subject: Re: The ||= assignment operator

On Apr 5, 12:08 pm, "David A. Black" <dbl...@rubypal.com> wrote:
> >> I think you mean:
>
> >>    lval || lval = rval
>
> > Huh?
>
> > My understanding has always been that x ||= y is shorthand for x = x
> > || y just as x += y is shorthand for x = x + y. That also seems to be
> > the understanding of the "Programming Ruby" text (p. 125 of the
> > latest). In other words, an assignment will take place even if it's
> > superfluous.
>
> Strictly speaking, it isn't shorthand for either, since there are
> cases where the expansion will fail but ||= won't because x isn't
> initialized. However, discounting that, the expansion is:
>
>    x || x = y
>
> The only time this matters is with hashes that have default values. In
> every other case, as far as I know, x = x || y also describes what's
> happening. But the expansion which describes *every* case is x || x =
> y.
>
> I wrote a blog post about this recently:http://dablog.rubypal.com/2008/3/25/a-short-circuit-edge-case
> (I changed || to or for some reason, which screws up the precedence,
> but I inserted corrections later.)

Interesting - nice blog post. That is surprising, inconsistent and
unfortunate :(