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Re: Changing variable to numeric without assignment (was:Using?smart matching)

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From:
Abigail
Date:
January 27, 2011 03:23
Subject:
Re: Changing variable to numeric without assignment (was:Using?smart matching)
Message ID:
20110127112345.GW8853@almanda
On Thu, Jan 27, 2011 at 11:17:18AM +0000, Ed Avis wrote:
> Abigail <abigail <at> abigail.be> writes:
> 
> >>>>>   $ perl -E '$a = $b = "abc100"; 0 + $b; say $a; say $b;
> >>>>>              $a ++; $b ++; say $a; say $b'
> 
> >>just evaluating the expression ($b + 0) in some other code, even if you
> >>do not assign the result to $b or try to change $b in any way, has the
> >>unexpected side-effect of changing the value stored in $b.
> 
> >Not changing the value stored in $b means that
> >
> >    my ($a) = /([0-9]+)/;
> >    @foo = map {$_ + $a} @bar;
> >
> >becomes more expensive, and $a will need to be converted from string to
> >integer for each element of @bar, instead of doing it once and caching
> >the result.
> 
> I see.  Personally, I'd take 'correct, but slightly slower' over 'a bit faster
> in some cases, with surprising edge cases' any day.

That's why you should never use floating point numbers, right? ;-)

> have it both ways?  The numeric value could be cached in the scalar and used
> next time, but without flipping the switch to mark the scalar as numeric.  Then
> later numeric context uses would be fast, but the value would not mysteriously
> change.
> 
> I have only a loose idea of Perl internals so please forgive my ignorance here;
> from a user's point of view not changing a variable just because it was *read*
> in an arithmetic expression is definitely the right thing to do, but I am
> uncertain of the practicalities of doing so in the perl interpreter.


That would probably mean we have to use a bit (or two, or three, if we want
to use the same trick for floats and strings) in the very limited supply
of bits reserved in the SV structure, for a case that's only seldom useful.



Abigail

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