This week's book giveaway is in the Mac OS forum. We're giving away four copies of a choice of "Take Control of Upgrading to Yosemite" or "Take Control of Automating Your Mac" and have Joe Kissell on-line! See this thread for details.

x = -1 y= x++ + ++x operands are always evaluated from left to right before an operation is performed all the operands are evaluated except in conditional operators such as (&& , || and in some cases of ? then the operation is performed based on precedence ans associativity. before addition could be performed both the operand has to be evaluated Left operand x++ is evaluated . since it is postfix expression the vlaue of the expression is the value of x before the increment is performed. so the expression will get the value -1(which is the value of x before increment) but x will be incremented by one so x will be -1+1=0. Right operand ++x is evaluated . since it is prefixx expression the value of the expression is the value of x after the increment is performed. so the expression will get the incremented value of x that is 1. Left expression + Right expression -1 + 1 = 0 so y=0 x =1 hope this helps Saravanakumar R

Hello Alexan: First the postfix and prefix operators are evaluated and the added together as they have higher precedence over addition. So coming back to your question: int x = -1; y = x++ + ++x; This would be evaluated as: y = -1 + 1 y = 0 The postfix operator (x++) gets the value -1 first and then x is incremented by 1 to give 0. The prefix operator (++x) adds 1 to x first, then uses the new value of x as the value of the expresiion. Hope this helps. Emad

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