YAAO (yet another assignment operator)

[Jack Jansen]

Dean Rubine asked for a ->= operator, but this
is quite impossible (if you try to keep the language
clean, of course). -> isn't an operator, it's a syntactical
construct, like . or [expr].
Also, it would be awfull to implement. In a statement
   p ->= next
you would have to keep the type of 'p' around, and also the
expression on the righthandside would not be an ordinary
expression, neither a lvalue nor a rvalue, but something wierd
(a 'selection expression'?).
I hope this isn't implemented. It doesn't do a lot of good,
and it's clearly a very dirty feature.
-- 
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