This is how Perl is interpreting the critical line of your code:

C:\test>perl -MO=Deparse,p -e" print (shift @a), (shift @a), "\n";"
print(shift @a), shift @a, \'n';
-e syntax OK
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As you can see, only the first parameter is being passed to print.

If you add another set of parens:

print( (shift @a), (shift @a), "\n" );
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Or drop the parens entirely:

print shift @a, shift @a, "\n";
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Your code will (probably) work as you expect.

Thank you, you are correct! Re-examining my code in the interim between posting and your timely response, I noticed that I was in fact receiving a warning:

print (...) interpreted as function at - line 7.
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Thank you for the clear explanation which has rectified my misunderstanding! I will also examine further your use of -MO=Deparse,p -e, which looks to be very useful indeed.

When you get a message like "print (...) interpreted as function ...", you can typically disambiguate your statement for Perl by simply adding a + in front of the opening parenthesis.

Your original code runs like this:

ken@ganymede: ~/local/bin
$ perl -Mstrict -Mwarnings -E 'my $s = "foo bar baz qux 3 3 1 3"; my @
+a = split /\s/, $s; splice(@a, 0, 4); while (@a) { print (shift @a), 
+(shift @a), "\n"; }'
print (...) interpreted as function at -e line 1.
Useless use of a constant (
) in void context at -e line 1.
31ken@ganymede: ~/local/bin
$
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Adding a single + (changing ... print (... to ... print +(...), you get the output you were after:

ken@ganymede: ~/local/bin
$ perl -Mstrict -Mwarnings -E 'my $s = "foo bar baz qux 3 3 1 3"; my @
+a = split /\s/, $s; splice(@a, 0, 4); while (@a) { print +(shift @a),
+ (shift @a), "\n"; }'
33
13
ken@ganymede: ~/local/bin
$
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This is discussed in more detail in perlop - Symbolic Unary Operators.

-- Ken

which looks to be very useful indeed.

Its so useful, its part of the Basic debugging checklist Tutorial.

Do note you only get said warning because you use exactly one space between print and the parenthesis. Drop the space, use 2 spaces, or a tab, and the warning disappears. Replace print by warn, and the warning disappears as well.

I see you got your answer as to what was happening in your code snippet from BrowserUk.

I don't know what context you want to use this in but the pairwise() function in List::MoreUtils may be helpful.

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.

... 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
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... 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
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