unary operator

[sanpan]

hi,
everybody
i have some confusion while working on post++ operater.
output of following code results 0,while according to
me it should be 2.
int i=0;
i=i++;
i=i++;
i=i++;
because after first line i should be 1,and after second
statement is executed i should be two;
please explain;

[bill bozeman]

This has been talked about many times in this forum. Do a search for i=i++ to get some great discussions on it. But basically it has to do with assignment presedence and since assignment has the lowest presedence you are getting something like i = 1 = 0, so you go backwards and i ends up as 0 again. I don't think that is the technical way of putting it, but that is how I remember it.
Bill

[sanpan]

thanks bill
but sorry i am not satisfied with your answer.
you say
i=0,i=i++
will convert into i=1=0 but i think since on i post increment
operator is applied so it should be first evaluvated then
incremented . so it should be i=0=1;
please help me .

thanks in advance

[Jane Griscti]

Hi Sanpan
Here's what happens with postfix operator:
<pre class="code">
int x = 0;
x = x++; <span class="red">// result: 0, x is not incremented</span>

original value of x is saved (x^0rig)
x is incremented
x^0rig is assigned to x
therefore, x will always equal original value
</pre>
Hope that helps.
Moving this thread to Certification Study.
------------------
Jane Griscti
Sun Certified Java 2 Programmer
"When ideas fail, words come in very handy" -- Goethe

[Chee]

Hi
Here you need to understand the concept of assignment operator.
Everytime you are trying to assign the post - incremented value of i++ to i the value is never assigned..Hence you get 0.
i=0;
i=i++; Here the value of i is 0 Since i++ is not incremented.
i=i++; Though previous i++ has increased to 1 it's not assigned to i Hence still i remains 0. Same is the case for all the 3 lines..
This explains you query..
Cheers Johnson