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Home/ Questions/Q 8704991
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Editorial Team
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Editorial Team
Asked: June 13, 20262026-06-13T03:14:37+00:00 2026-06-13T03:14:37+00:00

The following snippet is not working as expected: $k{foo}=1; $k{bar}=2; if(not defined($k{foo}) && not

  • 0

The following snippet is not working as expected:

$k{"foo"}=1;
$k{"bar"}=2; 
if(not defined($k{"foo"}) && not defined($k{"bar"})){
    print "Not defined\n";
}
else{
    print "Defined"
}

Since both $k{“foo”} and $k{“bar”} are defined, the expected output is “Defined”. Running the code, however, returns “Not defined”.

Now, playing around with the code I realized that placing parentheses around each of the not defined() calls produces the desired result:

if((not defined($k{"foo"})) && (not defined($k{"bar"}))){print "Not Defined"}

I imagine this has something to do with operator precedence but could someone explain what exactly is going on?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-13T03:14:38+00:00Added an answer on June 13, 2026 at 3:14 am

    Precedence problem.

    not defined($k{"foo"}) && not defined($k{"bar"})
    

    means

    not ( defined($k{"foo"}) && not defined($k{"bar"}) )
    

    which is equilvalent to

    !defined($k{"foo"}) || defined($k{"bar"})
    

    when you actually want

    !defined($k{"foo"}) && !defined($k{"bar"})
    

    Solutions:

    • !defined($k{"foo"}) && !defined($k{"bar"})
    • not defined($k{"foo"}) and not defined($k{"bar"})
    • (not defined($k{"foo"})) && (not defined($k{"bar"}))

    PS – The language is named “Perl”, not “PERL”.

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