in reply to arrays: shifting in while loop

My understanding is that the while will execute so long as there are elements in @a...

ivanthemad: I expect you already understand the following point perfectly well, but I want to dispell my sneaking suspicion of a possible misapprehension on your part.

The while-loop in the example code will only begin execution (Update: of a loopthe loop body) if the array is non-empty. However, (Update: once begun) looploop body execution will continue regardless of the state of the array unless an explicit loop exit is made conditional upon array state, e.g., with a statement like  last unless @a;.

Update: Clarified (hopefully) pertinence to loop body per repellent.

Replies are listed 'Best First'.
Re^2: arrays: shifting in while loop
by repellent (Priest) on Mar 07, 2012 at 05:01 UTC
      ... loop execution will continue regardless of the state of the array unless an explicit loop exit is made ...

    Hmm, really? I always thought the while-loop EXPR was re-evaluated after each iteration:
    my @a = (3, 4);

while (do { say "yo"; @a })
{
  say "hi mom";
  shift @a;  # no explicit loop break out
}

say "bye mom";

__END__
yo
hi mom
yo
hi mom
yo
bye mom

    [download]
[reply]
[d/l]
      ... the while-loop EXPR [is] re-evaluated after each iteration...

      It is, but I was trying to make a (perhaps rather trivial) point with regard to continued execution of the loop body (which I've tried to clarify in the original reply) as exemplified below. 

      >perl -wMstrict -le
"my @ra = (9, 8, 7);
 ;;
 while (@ra) {
   print 0+@ra, ' in @ra';
   shift @ra;
   print 0+@ra, ' in @ra';
   shift @ra;
   print 0+@ra, ' in @ra';
   shift @ra;
   print 0+@ra, ' in @ra';
   shift @ra;
   print 0+@ra, ' in @ra';
   shift @ra;
   print 0+@ra, ' in @ra';
   }
"
3 in @ra
2 in @ra
1 in @ra
0 in @ra
0 in @ra
0 in @ra

      [download]
[reply]
[d/l]
        Great, I knew I missed something from your original post.
[reply]

In Section Seekers of Perl Wisdom