by analogy with += 
$foo ~= s/a/b/;

[download]
should be equivalent to 
$foo = $foo ~ s/a/b/;

[download]
so maybe ~ should be a binary operator that returns the result of the substitution whereas ~= would apply the substitution to $foo

In reply to Re: Re: regexp operator -- same mistake over and over again by Anonymous Monk
in thread regexp operator -- same mistake over and over again by rkg

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