in reply to Re: Hash of hashes syntax
in thread Hash of hashes syntax

I'm afraid your last solution, as concise as it might be, simply does not work. \%h always refers to the same hash. You may well change its content at each loop, you're always modifying the same hash, and all members of %hash_of_hashes refer to it.

You really must create a new hash at each loop. This will do it:

$hash_of_hashes{$rn} = { %h };

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(the hash keys and values are flattened into a list, which in turn is used to create an anonymous hash).

--bwana147

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Re: Re: Re: Hash of hashes syntax
by Enlil (Parson) on Mar 19, 2003 at 09:19 UTC
    I'm afraid your last solution, as concise as it might be, simply does not work. \%h always refers to the same hash

    Not true.

    In this case, it would refer to a different hash. This is because the variable my %h is lexically scoped, and thus it is recreated each time through the loop, and so \%h will point to a different hash each time through the loop. To illustrate what I mean here is some code:

    Read more... (6 kB)
    The differences are subtle, but jdporter's code works.

    -enlil

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Re: Re: Re: Hash of hashes syntax
by Chmrr (Vicar) on Mar 19, 2003 at 09:24 UTC

    Good eye, but this actually isn't a problem in this case. Because of the my %h inside of the loop, %h is a different (and new) lexical through each iteration -- thus there is no problem just sticking a reference to it into the hash or array. The problem comes from code like this (nearly directly copied from perldsc):

    my %hash;
for $i (1..10) {
  %hash = somefunc($i);
  $hash_of_hashes{$rn} = \%hash;     # WRONG!
}

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    Note where the "my" is. See perldsc or perlreftut or perllol for more.

    perl -pe '"I lo*`+$^X$\"$]!$/"=~m%(.*)%s;$_=$1;y^`+*^e v^#$&V"+@( NO CARRIER'

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