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


sub process_variables {
    my ( $value, $remove_undef ) = @_;

    $value =~ s/ (?<! \\ ) \$ \{ ( .*? ) \} /
        defined $variable{ $1 }
            ? $variable{ $1 }
            : defined $remove_undef && $remove_undef
                # for if statements, any variable that is not defined,
                # we simple convert to 0
                ? '0'
                # This could be an option that is used later, save
                # it so we don't warn if this option is not one of
                # ktests options. And put back the origin piece.
                : ( $used_options{ $1 } = 1 and "\${$1}" )
        /exg;

    return $value;
    }

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The code above produces the warning message:
Found = in conditional, should be == at file line n.
If I change:
( $used_options{ $1 } = 1 and "\${$1}" )
To:
do { $used_options{ $1 } = 1 and "\${$1}" }
I still get the same warning message.

But obviously the message is wrong because there is no = in conditional. 

$ perl -v

This is perl 5, version 24, subversion 1 (v5.24.1) built for x86_64-li
+nux-gnu-thread-multi
(with 90 registered patches, see perl -V for more detail)

Copyright 1987-2017, Larry Wall

Perl may be copied only under the terms of either the Artistic License
+ or the
GNU General Public License, which may be found in the Perl 5 source ki
+t.

Complete documentation for Perl, including FAQ lists, should be found 
+on
this system using "man perl" or "perldoc perl".  If you have access to
+ the
Internet, point your browser at http://www.perl.org/, the Perl Home Pa
+ge.

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Replies are listed 'Best First'.
Re: warning! -- Why?
by haj (Vicar) on May 12, 2022 at 07:05 UTC

    The and is the conditional here. The equal sign = has a higher precendence than and and Perl guesses (it's a warning, not a syntax error) that you meant to have a comparison on its left side. The expression might also not do what you want since the result of and is not its second operand, but I guess you want "\${$1}" as replacement string.

    The cleanest (IMO) way to get rid of that warning is to use do, as you did, but to replace the and in the do block with a semicolon:

    do { $used_options{ $1 } = 1;  "\${$1}" }

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      Bingo!

      Yes, I figured this out a couple of hours after I posted.

      Thanks

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Re: warning! -- Why?
by Athanasius (Archbishop) on May 12, 2022 at 07:14 UTC
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      Thanks. It is also about 15% faster than using do { ... } !

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Re: warning! -- Why?
by Tux (Canon) on May 12, 2022 at 06:59 UTC
    ( $used_options{ $1 } = 1 and "\${$1}" )

    is a condition. expression and action could also be written as if (expression) { action } or action if expression and in that expression, your = should clearly be ==.


    Enjoy, Have FUN! H.Merijn
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Re: warning! -- Why?
by hippo (Archbishop) on May 12, 2022 at 08:54 UTC
    ( $used_options{ $1 } = 1 and "\${$1}" )

    I fail to see why you are using and here in the first place. Is there any scenario under which $used_options{ $1 } = 1 evaluates to false?

    🦛

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Re: warning! -- Why?
by kcott (Archbishop) on May 12, 2022 at 09:37 UTC

    G'day jwkrahn,

    It's a common mistake to forget, or not realise, that "&&" has a higher precedence than "=" which, in turn, has a higher precedence than "and". See "perlop: Operator Precedence and Associativity" (there's a table at the end of that section which lists the order of precedence).

    Changing "and" to "&&" gets rid of the warning; however, in this instance, that's probably not what you want. Compare these two: 

    $ perl -MO=Deparse,-p -e '($used_options{ $1 } = 1 and "\${$1}")'
(($used_options{$1} = 1) and ("\${$1}"));

$ perl -MO=Deparse,-p -e '($used_options{ $1 } = 1 && "\${$1}")'
($used_options{$1} = "\${$1}");

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    — Ken

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