use Perl

• Are these equal? (Score:2, Interesting)

  by $Bob (668)

  Looking at An Aside: the "Smart Match" Operator on page 3, are these equal?
  
  ($elem =~ @array) == (@array =~ $elem)
  
  ($key =~ %hash) == (%hash =~ $key)
  
  ($value =~ (1..10)) == ((1..10) =~ $value)
  
  ($value =~ ('a', /\s/, 7)) == (('a', /\s/, 7) =~ $value)
  
  • Re:Are these equal? (Score:3, Informative)

    by TimToady (2729)

    Yes.
    
    Larry
    • Re:Are these equal? (Score:1)

      by hossman (2110) on 2002.04.06 15:34 (#6697)

      Are you saying that the =~ is going to be defined such that ($paco =~ $yakko) == ($paco =~ $yakko) will allways be true? (or are you saying that just the 4 examples $Bob listed will allways be true)

      What if $paco and $yakko are both sub refs? there's no garuntee that $yakko.($paco) will ever be equal to $paco.($yakko).

      for that matter, what about the list behavior Damian describes...

      if $value =~ ('a',/\s/,7) {...}
      # true if $value is eq to 'a'
      # or if $value contains whitespace
      # or if $value is == to 7
      

      That final example illustrates some of the extra intelligence that Perl 6's =~ has: When one of its arguments is a list (not an array), the "smart match" operator recursively "smart matches" each element and ORs the results together, short-circuiting if possible.

      What happens in this scenerio...

      if ($sub1 =~ ($sub2, $sub3)) { ... }

      is it equivilent to ($sub1.($sub2) || $sub1.($sub3)) ? Or ($sub2.($sub1) || $sub3.($sub1)) ?

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      • Re:Are these equal? (Score:2)

        by Damian (784)

        Are you saying that the =~ is going to be defined such that ($paco =~ $yakko) == ($paco =~ $yakko) will allways be true? (or are you saying that just the 4 examples $Bob listed will allways be true)

        He's saying the latter. The matching rules are supposed to be "sensible" and "reasonable" (for sufficiently vague values of those two words ;-)

        The table in A4 [perl.com] lists the behaviours of =~ under various combinations of arguments.

        What if $paco and $yakko are both sub refs? there's no garuntee that $yakk