From: Robert Klemme <shortcutter@...>
Date: 2007-09-04T22:37:26+09:00
Subject: Re: assigning to hash keys when there is a default value?

2007/9/4, Yossef Mendelssohn <ymendel@pobox.com>:
> On Sep 4, 7:32 am, "Robert Klemme" <shortcut...@googlemail.com> wrote:
> > 2007/9/4, Pe�a, Botp <b...@delmonte-phil.com>:
> >
> > > From: sco...@gmail.com [mailto:sco...@gmail.com] On Behalf Of Russell Norris:
> > > # I learned that x ||= y means set x to y unless x, so I don't see the
> >
> > > there's the bug: you've changed the meaning of x ||= y, or of foo<op>=bar for that matter.
> >
> > > # bug. and i don't see this as being at all unpredictable.
> >
> > > you just said you've _learned that x||=y means set x to y unless x. Surely it was _unpredictable at some point. Surely x<op>=y is a no-brainer for a many of us :)
> >
> > > and now we all have to update our test cases, cause maybe x += 1 may not always be x = x + 1
> >
> > > speaking of least surprise :(
> >
> > No, "x+=1" will always be "x=x+1".
> >
> > robert
>
> If '"x+=1" will always be "x=x+1"', what's the problem in having x ||=
> 1 always be x = x || 1?

That an assignment of x=x is useless (basically a nop in the standard
case of x being a variable).  Please see also one of my earlier
postings.

Kind regards

robert