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Re: Thoughts on replacing -> with .
in thread Thoughts on replacing -> with .
Note that =~ and ~= are both valid Perl already.
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?
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 ~=.
$a ~= $b
But XOR-assign is supposed to be $a ^= $b...
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