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in reply to Re: Thoughts on replacing -> with .
in thread Thoughts on replacing -> with .

Note that =~ and ~= are both valid Perl already.

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Re^3: Thoughts on replacing -> with .
by Ralesk (Pilgrim) on Jun 12, 2013 at 11:27 UTC

    Hmm, I don’t see ~= in perlop; and ~ is the unary bitwise negation operator, so it wouldn’t make much sense to have ~=. What is it?

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      It is like += and .=, which modify the left-hand side by the right hand side.

      $a += $b adds the right hand side ($b) to the left hand side, $a, and assigns the result to $a.

      $a .= $b appends $b to $a and assigns the result to $a.

      $a ~= $b xor's $b and $a and assigns the result to $a.

      As this is a relatively uncommon use, it maybe makes this interchange currently even harder to spot than later the use of =~ and ~=.

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        But XOR-assign is supposed to be $a ^= $b...

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        I am getting a syntax error for 
        $a ~= $b

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