in reply to Unexpected persistence with each hash

The seemingly sane version could use some strictness. my $key = each %val makes no sense. The slightly more sane my $key = each %foo still makes no sense. each returns a list. $key is now always 2, unless it's 0.

Oh - try seeing what the stringification of %foo is, even when it's empty. It's not what you think. (And, even though overload has a hook for stringy, nummy, and booly - I'm pretty positive that scalars, arrays, and hashes all use the same values for each hook.)

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Re: Re: Unexpected persistence with each hash
by demerphq (Chancellor) on Jul 01, 2003 at 07:53 UTC

    each returns a list. $key is now always 2, unless it's 0.

    Not so. each is context sensitive as are most things in Perl (you knew that, silly :-)

    each HASH
    When called in list context, returns a 2-element list consisting of the key and value for the next element of a hash, so that you can iterate over it. When called in scalar context, returns only the key for the next element in the hash.

    Actually there is an interesting point here, iirc when you do a Tie::Hash implementation the EACH() sub only returns the key. Perl then automatically follows with a FETCH() to get the value (and I think it does so regardless of the context as well that the each keyword is used in). This is a bit annoying as often in this situation you need to do the same work to get the key or the value and finding one gives you the other. (You can play games to optimise this but its still annoying.)

    Oh - try seeing what the stringification of %foo is, even when it's empty. It's not what you think.

    I would have thought it would be false when the hash is empty, and suprise suprise it is. :-)

    The only change Zaxo needed to make to his code was as you pointed out to change the %val to %foo. Your points about strictures etc is good of course :-)

    my %foo; @foo{qw/ foo bar baz /} = 1 .. 3; printf "%%foo = %s { %s } = ( %s )\n",\%foo,scalar %foo, join(",",map "$_ => $foo{$_}",keys %foo); while (%foo) { my $key = each %foo; my $val = delete $foo{$key}; printf "%s => %s is gone.\n", $key, $val; printf "%%foo = %s { %s } = ( %s )\n",\%foo,scalar %foo, join(",",map "$_ => $foo{$_}",keys %foo); }
    %foo = HASH(0x1ab298c) { 2/8 } = ( foo => 1,baz => 3,bar => 2 ) foo => 1 is gone. %foo = HASH(0x1ab298c) { 1/8 } = ( baz => 3,bar => 2 ) baz => 3 is gone. %foo = HASH(0x1ab298c) { 1/8 } = ( bar => 2 ) bar => 2 is gone. %foo = HASH(0x1ab298c) { 0 } = ( )

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    demerphq

    <Elian> And I do take a kind of perverse pleasure in having an OO assembly language...
Re: Re: Unexpected persistence with each hash
by Zaxo (Archbishop) on Jul 01, 2003 at 03:56 UTC

    Thanks, stricture would have indeed caught the typo. As you suggest, my $key = each %foo; was what I meant. The corrected code,

    my %foo; @foo{qw/foo bar baz/} = 1 .. 3; while (%foo) { my $key = each %foo; my $val = delete $foo($key}; printf "%s => %s is gone.\n", $key, $val; }
    works as intended.

    In scalar context, each returns only a key.

    /me claims the brown bag for today. What you refuse to see, is your worst trap in action. Thanks also to ++Albannach.

    After Compline,
    Zaxo