Re: eval question

Is it possible to do $r = eval("$a $op $b")

In principle yes, if you map the operators 'le', 'ge' etc. to the corresponding Perl equivalents — e.g. using a hash.

my %perlop = (
   le => '<=',
   ge => '>=',
   #...
);

my $r = eval "$a $perlop{$op} $b";
[download]

(But as always with string eval, be careful what you interpolate when the data ($a, $b here) doesn't come from trusted sources...)

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Re^2: eval question by moritz (Cardinal) on Jan 06, 2011 at 19:59 UTC
(But as always with string eval, be careful what you interpolate when the data doesn't come from trusted sources...) If you whitelist allowed operators via hash, and don't interpolate `$a` and `$b`, you should be fine: `if (exists $perlop{$op}) { my $r = eval "\$a $perlop{$op} \$b"; } else { die "OH NOEZ!"; }` [download] That way the string passed to eval contains the variable names, and obtains their value from the outer scope. Perl 6 - second systems done right	[reply] [d/l] [select]

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Re^2: eval question
by moritz (Cardinal) on Jan 06, 2011 at 19:59 UTC

(But as always with string eval, be careful what you interpolate when the data doesn't come from trusted sources...)

If you whitelist allowed operators via hash, and don't interpolate $a and $b, you should be fine:

if (exists $perlop{$op}) {
    my $r = eval "\$a $perlop{$op} \$b";
} else {
    die "OH NOEZ!";
}
[download]

That way the string passed to eval contains the variable names, and obtains their value from the outer scope.