assignment operators

[Anna Chu]

I get a very wierd problem about assigment operator about the code below:

int i=0;
i=i++;
System.out.println(i);

The result printed out is 0.

I don't know why it is not 1. Can someone explain how this happens?
Thanks a lot!

[Michael Carlson]

It's post increment, the assignment happens and then i is incremented.

[wise owen]

PostIncrementOperatorAndAssignment

[Ashok S Yadav]

I am confused with the reply.
int i =0;
i = i++;
System.out.println(i);
Here is what I think the sequence of operations is
1. i=0 and this value is assinged to the RHS and hence i is zero.
2. i is then incremented due to the post fix increment.
Shouldn't i then become 1.

Thanks

[Naseem Khan]

Posted by Michael Carlson
It's post increment, the assignment happens and then i is incremented.

Its a misconception among ppl that assignment occurs before increment but its wrong.


If u see the precedence order, postfix operator has higher precedence than assignment. So i++ is executed first.

int i=0;
i=i++;
System.out.println(i);

      (i=(i++));

Here is the order of execution.

1. i++ gets evaluated, so i becomes 1 but still the value of the expression i++ is zero.

U can test it by...

System.out.println(i++); it prints 0. if i is zero initially

       (i=([B]i++[/B]));
            [B]|[/B]
            [B]v[/B]
           ([B]0[/B])
        i = 0

Even in method call,

meth(i++, i) means-->>> meth(0, 1) if i is initially zero.

jst remember this. first VM stores the value of i which is zero to some temporary location then increments its value by one. So i becomes one, but the value which is assigned to i is again temporary location value.

Hence i again becomes zero.

Naseem
[ May 31, 2006: Message edited by: Naseem Khan ]