> ++a + a++
> what values actually get used as operands for the '+' and what the final
> resulting value of the variable are pretty bad mojo to sort out.
I would hope (wrongly again probably) that NOTHING would change; it would
be:
a = 1;
--a = 0 + a++ = 1 == 1
Which means that --a + a++ = what a was to start with.
PS - Sorry about r-l earlier -- I guess I should have said:
$result = &operation1 + &operation2;
Operation order shouldn't matter much...
Unless we have global vars...
???
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