Re^3: Thoughts on replacing -> with .

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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Re^4: Thoughts on replacing -> with . by Corion (Patriarch) on Jun 12, 2013 at 11:32 UTC
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 `~=`.	[reply] [d/l] [select]
Re^5: Thoughts on replacing -> with . by Ralesk (Pilgrim) on Jun 12, 2013 at 11:38 UTC
But XOR-assign is supposed to be `$a ^= $b`...	[reply] [d/l]
Re^5: Thoughts on replacing -> with . by choroba (Cardinal) on Jun 12, 2013 at 11:38 UTC
I am getting a syntax error for `$a ~= $b` [download] لսႽ† ᥲᥒ⚪⟊Ⴙᘓᖇ Ꮅᘓᖇ⎱ Ⴙᥲ𝇋ƙᘓᖇ	[reply] [d/l]
Re^6: Thoughts on replacing -> with . by Corion (Patriarch) on Jun 12, 2013 at 11:47 UTC
Err - yes. I wonder how I came to this interpretation and I'm sorry I did not test it before...	[reply]
Re^6: Thoughts on replacing -> with . by Grimy (Pilgrim) on Jun 13, 2013 at 18:21 UTC
Correct. On the other hand, `:$ ~= $b` does work, but is has nothing to do with XOR.	[reply] [d/l]

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