in reply to late binding behavior in hash definition

You might as well ask why the right-hand side of 
    $x = ($x + 18) / 7;

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does not take into account the value that you are trying to assign, before it is computed, and set $x to 3.

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Re: Re: late binding behavior in hash definition
by japhy (Canon) on Nov 22, 2000 at 19:38 UTC
[reply]
      I'm not kidding. Some languages have something a little like that. But you quickly get into big problems. Consider: 
          my $x = not $x;
    my $y = $y-1;

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      or 
          my @a = (@a, 3);

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      for example.
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