This week's book giveaway is in the OCAJP 8 forum. We're giving away four copies of OCA Java SE 8 Programmer I Study Guide and have Edward Finegan & Robert Liguori on-line! See this thread for details.

this output 5 as after ++y ,y become 2 then add to it y++ where y =2 in that case then 2+2 =4 then increment y to be 3 then add the left hand side operands of y to have y=y+4= 3+4= 7
??? that what i understand but i don't know how 5 is the output ??

also try to test the same code changing y to int y=0; and see the result that =2

SCJP Pass 85%....
Thousands of candles can be lighted from a single candle, and the life of the candle will not be shortened

this output 5 as after ++y ,y become 2 then add to it y++ where y =2 in that case then 2+2 =4 then increment y to be 3 then add the left hand side operands of y to have y=y+4= 3+4= 7

First of all try to simplyfy the expression.
Actual expression is

y= y+ ++y + y++;

Here y=1. And when you say y+ that is 1+
then it comes to ++y here y becomes 2

expression is y= 1 + 2 + y++ ok?

then it becomes like this y=1 + 2 + 2 (And not 3 Because it is post increment operator so it first uses current value then
increment it.)

Thats why... finally y= 1 + 2 + 2 i.e. 5

In case of 0.... I think you can follow the above steps.
You'll get the answer as 2.

1) Precedence Order: When two operators share an operand, the operator with higher precedence goes first.. I think, you know that!
2) Associativity : When two operators with the same precedence the expression is evaluated according to its associativity.
For Ex :

is treated as

Since = operator has right to left associativity.

3) Order of evaluation : In Java, the left operand is always evaluated before the right operand. also it applies to function arguments.