http://qs321.pair.com?node_id=138034

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

why does this not work? 
my value = 444;

if ($value == 456 || 453 )
{
print something;
}
else 
{
print nothing;
}

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Why is this printing something rather than nothing.

peace, LOVE and ((code))

basicdez

Replies are listed 'Best First'.
Re: Help with this statement
by jwest (Friar) on Jan 11, 2002 at 19:59 UTC
    Your if clause looks to perl like:
    if (($value == 456) || (453))

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    instead of what you intend.

    --jwest

    -><- -><- -><- -><- -><-
All things are Perfect
    To every last Flaw
    And bound in accord
         With Eris's Law
 - HBT; The Book of Advice, 1:7
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Re: Help with this statement
by talexb (Chancellor) on Jan 11, 2002 at 20:29 UTC
    The statement 
    if ($value == 456 || 453 )

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    works out as follows: "If variable value equals 456 OR if 453 is non-zero ...". Of course, 453 will always be non-zero. :)

    What you probably want to say is 

    if ($value == 456 || $value == 453)

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    --t. alex

    "Excellent. Release the hounds." -- Monty Burns.

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Re: Help with this statement
by tradez (Pilgrim) on Jan 11, 2002 at 20:48 UTC
    I believe your problem is the non-transative nature of the perl || and 'or' operators. in this case you are asking if it is equal to 456 or just 453. 453 of course is never 0 and therefore will be true. Hence printing something. to fix this just add. 
    if ($value == 456 || $value == 453){
print something;
}
else 
{
print nothing;
}

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    That should clear things up.
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[d/l]
      I couldn't resist this as a TIMTOWTDI :) 
      unless (abs($value - 454.5) - 1.5) {
  print something;
}
else 
{
  print nothing;
}

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      cLive ;-)

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[d/l]
        just as an alternative, it can also be written as: 
        my $value = 444;

if ($value == 456) { print "something"; }
elsif ($value == 453) { print "something"; }
else { print "nothing"; }

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[reply]
[d/l]
      Thanks I got it working now. peace dez L
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Re: Help with this statement
by BazB (Priest) on Jan 11, 2002 at 20:02 UTC
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(crazyinsomniac) Re: Help with this statement
by crazyinsomniac (Prior) on Jan 12, 2002 at 09:01 UTC
    If you can't memorize perlop and the precedence, and I tell you I cannot, the way to figure out how "perl" is reading yer program, is to use O::Deparse, so that's what I did (although this example was simple enough to figure out): 
    bash-2.05$ perl -e'$v = 444;if($v == 456 || 453){print 1}else{print 2}
+'
1bash-2.05$
bash-2.05$
bash-2.05$
bash-2.05$ perl -MO=Deparse -e'$v = 444;if($v == 456 || 453){print 1}e
+lse{print 2}'
-e syntax OK
$v = 444;
if ($v == 456 or 453) {
    print 1;
}
else {
    print 2;
}

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    Now that's not what you thought would happen. Perl is pretty cool, and there is much voodoo within, but the *binary* operators, cannot be chained as such . The term binary implies that they operate on 2 values, the one on the left, and the one on the right. it's not the one on the left, and the 2 on the right (usually). ++ and -- unary operators. they operate on 1 value. now you know. peace

     
    ______crazyinsomniac_____________________________
    Of all the things I've lost, I miss my mind the most.
    perl -e "$q=$_;map({chr unpack qq;H*;,$_}split(q;;,q*H*));print;$q/$q;"

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Re: Help with this statement
by screamingeagle (Curate) on Jan 12, 2002 at 05:22 UTC
    Another way of doing this is simulate a switch-case construct.Here's an example from the Perl documentation on how to do this : 
    SWITCH: for (ref $whatchamacallit) {

    /^$/        && die "not a reference";

    /SCALAR/    && do {
                print_scalar($$ref);
                last SWITCH;
            };

    /ARRAY/        && do {
                print_array(@$ref);
                last SWITCH;
            };

    /HASH/        && do {
                print_hash(%$ref);
                last SWITCH;
            };

    /CODE/        && do {
                warn "can't print function ref";
                last SWITCH;
            };

    # DEFAULT

    warn "User defined type skipped";

    }

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