Re: foreach loop question

Why the following code does not process elements added inside the loop ...

Consider that

  foreach ( sort keys %h ) {
      ...
      $h{$new} = $new;
[download]

is equivalent to

  @sorted = sort keys %h;
  foreach ( @sorted ) {
      ...
     $h{$new} = $new;
[download]

The keys have been extracted from the hash before being sorted and processed. Any keys you add after that aren't "visible" to the foreach.

By the way, the "next;" is redundant.

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Re: Re: foreach loop question by dda (Friar) on Apr 16, 2003 at 18:45 UTC
Well, thanks. But without "sort" it works the same way. I wonder how it processes large arrays - always creates a copy? --dda	[reply]
Re: Re: Re: foreach loop question by dws (Chancellor) on Apr 16, 2003 at 18:54 UTC
Well, thanks. But without "sort" it works the same way. I should have been more explicit. It's "keys" that creates the new array, not the sort. If you want to process all of the keys and values in a hash without creating a new array to hold the hash keys, look into "each". `while ( ($key, $value) = each %h ) { ...` [download] This doesn't create an additional array. Note, though, that it's unsafe to manipulate the hash while you're iterating over it. The proviso in the docs reads: If you add or delete elements of a hash while you’re iterating over it, you may get entries skipped or duplicated, so don’t. Exception: It is always safe to delete the item most recently returned by "each()".	[reply] [d/l]