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Weird syntax question

by Ovid (Cardinal)
on Aug 23, 2001 at 02:23 UTC ( #107171=perlquestion: print w/replies, xml ) Need Help??

Ovid has asked for the wisdom of the Perl Monks concerning the following question:

I was trying to patch some buggy code when I came across some syntax that I thought was a bit odd and certainly doesn't return results that I expect. Here's the smallest test case I could replicate:

perl -e '$x=1;print 1 if $x or print 0'

That prints 1, just as I would expect, despite the syntax being a bit odd. However, if I set $x to a false value, it prints 01. Why the heck does it print the zero and then turn around and print the one, anyway? Since print, in this context, should always return a true value, shouldn't this short-circuit and not do both prints? And wouldn't the first print be skipped by the "if $x" check?

If I change this slightly:

my $x = 0; &one if $x or &zero; sub one { print "one"; 1; } sub zero { print "zero"; 0; }

That only prints "zero". I have an explicit return of false in &zero, but all in all, I'm just not quite grokking this.

Cheers,
Ovid

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Replies are listed 'Best First'.
Re: Weird syntax question
by John M. Dlugosz (Monsignor) on Aug 23, 2001 at 02:37 UTC
    The condition of the if is performed first: $x is false, so it needs the Right hand side of the OR. That prints 0, returns true, and then the if is satisfied (false or true => true). Then it prints 1 because the if said so.

    This implies that the or applies to the condition after the if, not to the statement including the if:

    print 1 if ($x || print 0)
    —John

      Good point. Both operands to an "or" are expressions, not statements, so you can't parse "print 1 if $x or print 0" as "( print 1 if $x ) or print 0" since "print 1 if $x" is not a valid expression (since "if" is a statement modifier).

              - tye (but my friends call me "Tye")
Re: Weird syntax question
by htoug (Deacon) on Aug 23, 2001 at 09:45 UTC
    Whenever you are confused by some wierd perl syntax alway run to the wonderfull B::Deparse module.

    If you use it with the -p option to add extra parentheses it will often (almost always) be clear what is going on.

    Using it on your example I get:

    perl -MO=Deparse,-p -e '$x=1;print 1 if $x or print 0' ($x = 1); (($x || print(0)) and print(1)); -e syntax OK
    It should be clear that all the explanations concerning 'or' and expressions are true.

    Your other question is related.
    The interesting part is the line &one if $x or &zero;B::Deparse deparses it as (($x || &zero) and &one); $x is false, so it must evaluate &zero.
    &zero prints 'zero' and returns false.
    So the entire expression is false and &one is not called.

    To sum it all up: use B::Deparse and all will be revealed.

Re: Weird syntax question
by Zaxo (Archbishop) on Aug 23, 2001 at 02:38 UTC

    print always succeeds (for a suitable definition of 'always'), i.e. print 0; returns 1.

    $ perl -e 'print print 0' 01$

    Update: I should also say that the short-circuit behavior of or omits the print 0 clause when $x is true.

    After Compline,
    Zaxo

      perl -e 'print qq(Gotcha!\n) unless print STDIN qq(Hello)'
      All interactions with the system have the possibility of failure...
Re: Weird syntax question
by rchiav (Deacon) on Aug 23, 2001 at 02:40 UTC
    Through a little poking around, this evaluates as..
    perl -e '$x=1;print 1 if ($x or print 0)'
    Which evalutates  ($x or print 0) first. That will always return true so the 1 gets printed in either case.

    With the subs, it will evaluate to false since you're returning 0.

    Rich

(tye)Re: Weird syntax question
by tye (Sage) on Aug 23, 2001 at 02:39 UTC

    "or print 0" is true unless the print fails so your if ends up being true as well which causes the "print 1" to be executed.

            - tye (but my friends call me "Tye")
Re: Weird syntax question
by flgr (Sexton) on Aug 23, 2001 at 14:19 UTC
    I'd probally do it that way: perl -e '$x=1;print $x'. But that only works with your (maybe cutted) example. If you want to get one out of two strings depending on the value of something use perl -e '$x=1;print $x?'foo':'bar'.

    $=lc(sub{}^" \6 \n");$=~s/\(.*\)//;for(split
    ('',$)){$/=$_;for(5..7){$/++}$:.=$/}print$:;1
Re: Weird syntax question
by Anonymous Monk on Aug 23, 2001 at 22:40 UTC
    What you have in mind is a re-arrangement of:

       if ( $x ) { print 1; } else { print 0; }

    What you are really getting is:

       if ( $x or print 0 ) { print 1; }
    

    Thus when $x is false, 'print 0' is executed, and then, since the print() returns true, the if-then body is carried out.

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