post and pre increment opeartor

[Heba Mahmoud]

  

int y =1;

y+ = ++y +y++;

  

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

[Aditya Kanitkar]

Heba Mahmoud wrote:

  

int y =1;

y+ = ++y +y++;

  

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.

[Abimaran Kugathasan]

Correct!

[Heba Mahmoud]

why you evalute the expression as

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

and didn't evalute it as the rule saying evalute the right operand first then the left operand

i understand what you do but i don't understand why not you evalute it as i made ??

[Abimaran Kugathasan]

Heba Mahmoud wrote:why you evalute the expression as

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

and didn't evalute it as the rule saying evalute the right operand first then the left operand

i understand what you do but i don't understand why not you evalute it as i made ??

There is a order.

In Java the Order of Evaluation is from left to right...

[Abimaran Kugathasan]

There are 3 things related to Operators in Java

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 :

x = y = z = 17;

is treated as

x = (y = (z = 17));

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.

For Ex :

int x = 5;
int z = ++x * x * x--; // Prints 216

Hope you understand this!

[Seetharaman Venkatasamy]

Heba Mahmoud wrote:

  
int y =1;
y+ = ++y +y++;

is nothing but y = y + (++y + y++);

Always use decompiler for this kind of code .

decompiled code is below:

    
int y = 1;
  y = y++ + (y + y++);
  System.out.println(y);