It is evaluated left to right: 1. y = x++ (At this point y = -1 and x now equals 0) 2. y = y + ++x (Now y = 0 and x = 1)

Any intelligent fool can make things bigger, more complex, and more violent. It takes a touch of genius - and a lot of courage - to move in the opposite direction. - Ernst F. Schumacher

Just a guess, I think here's what happened: x=-1 y=x++ + ++ x; // x= -1 x++ has the most precedence, so it becomes: y= (-1) + ++x; // x = 0 ++x is next: y = (-1) + (1); therefore the result is 0. It is interesting to compare this with: x = -1; x += x++; Which will give -2, in my mind the += operator works like: int tmp = x; x = tmp + x++; Hope this helps and the above is correct... Adrian

SCJP 1.4, SCWCD 1.4, SCBCD 1.3

Bojan Knezovic
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Originally posted by Michael Morris: It is evaluated left to right: 1. y = x++ (At this point y = -1 and x now equals 0) 2. y = y + ++x (Now y = 0 and x = 1)

I thought x gets inceremented only AFTER the = operator (for #1, x++), isn't that what's pre and postfix notation all about... Oh well... :roll: I thank you and Adrian.