> ++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... ??? _Sx____________________ ('> -Sx- IUDICIUM //\ Have Computer - v_/_ Will Hack... "iudicium ferat"