Operator/Regex question

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

I'm relatively new to writing Perl (though I've done my best to educate myself) and I've been tasked with reverse engineering some code. What is the purpose of the tilde (~) operator at the beginning of the following line of code? (this isn't the binding pattern match operator =~)

~s/(pattern)/replacement/g;
[download]

Much thanks for your help. -Migloth

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Re: Operator/Regex question by ikegami (Patriarch) on Jun 03, 2009 at 16:10 UTC
Could it have been part of a "`=~`" token? Bitwise negation ("`~`") doesn't make much sense there, but "`=~`" binds the result of an expression to `m//`, `s///`, `tr//` and similar operators. Those operators will then match against that value, and possibly modify it. If no value is bound to the operator, they use `$_`. `my ($key, $val) = ( $pair =~ /^([^=]+)=(.*)/s );` [download] (The parens optional since "`=~`" has higher precedence than "`=`".) See perlop	[reply] [d/l] [select]
Re: Operator/Regex question by JavaFan (Canon) on Jun 03, 2009 at 15:59 UTC
Without context, it would be guessing. My guess it's a typo, going unnoticed because the return value isn't used.	[reply]
Re: Operator/Regex question by kennethk (Abbot) on Jun 03, 2009 at 15:59 UTC
As described in perlop, ~ is the bitwise negation operator. A substitution returns the number of substitutions performed, though I can't imagine a good reason why you'd want bitwise negation of that return value. Update: After reading JavaFan's comment below, I think you're looking at a victim of refactoring where no bugs were introduced.	[reply]
Re: Operator/Regex question by Anonymous Monk on Jun 03, 2009 at 15:59 UTC
perlop Unary "~" performs bitwise negation, i.e., 1's complement. `C:\>perl -e"print ~1" 4294967294 C:\>perl -e"print ~~1" 1` [download]	[reply] [d/l]