Assigning and Post incrementing:

[Narasimha Rao B.]

Hi,
Please clarify me, whether the below understanding of mine is correct or not?
int k = 3;
for(int i=0;i<5;i++){
System.out.println(k=k++)}
Above print statement will always print 3 and k value will not be incremented.
Reason is,
Initially k value is 3 and will be assigned to k, then k value will be incremented temporarily by one, i.e., now temporary k value is 4 and now I am not assigning the temporary value of k to any thing, hence it will lost and now the actual k value is 3, which I assigned before incrementing. Is this understanding is correct or not? If it is wrong what is actual logic behind this?
Thanks in Advance,
Narasimha.

[Vicken Karaoghlanian]

Forget the loop and let us focus on
int k = 3;
k = k++;
The JLS states that "The post increment operator returns its old value then increment the result"
Knowing this fact you can easily understand what happens:

'k' is initialized to the value of 3

'k++' is executed, thus k becomes 4 but it returns its old value which is 3

3 is assigned to 'k'

Remember that the increment operator has a higher priority over the assignment; therefore it is executed first (not the other way around as you suggested).
[ February 07, 2004: Message edited by: Vicken Karaoghlanian ]

[Davy Kelly]

hey narasimha,
I like to look at it in a similar fashion but with a slightly different method.
int k = 3;
k = k++; //k is assigned to 3 here
//but would be 4 here. because you have exited the loop, it does not get to be 4!
so every time you see k = k++;in your example it is saying, k = 3
Davy

[Vicken Karaoghlanian]

Originally posted by Davy Kelly:
hey narasimha,
I like to look at it in a similar fashion but with a slightly different method.
int k = 3;
k = k++; //k is assigned to 3 here
//but would be 4 here. because you have exited the loop, it does not get to be 4!
so every time you see k = k++;in your example it is saying, k = 3
Davy

Davy, always remember the fact that the increment operator has a higher priority over the assignment; therefore it is executed first.
How do you think the following expression is evaluated, and what is it output?
int k = 0;
k = k++ + ++k + k++;
Don't ask the compiler for the answer, try doing it yourself.

And by the way...

I no my spelling is carp!!!

I speaking english very best too.
[ February 07, 2004: Message edited by: Vicken Karaoghlanian ]

[Davy Kelly]

Vicken,

int k = 0;
k = k++ + ++k + k++;
as I see it is :
k++ is 0 (at the moment add 1 at end of statement)
+
++k is 1
+
k++ is 1 (because of last preincrement ++k but add 1 at end of statement)
so k = 2 (but have to add another 2 because we at end of statement.
k = 4
Can I compile it now to see if I am correct???
Davy

[Vicken Karaoghlanian]

Davy, your output is correct, but you missed the evaluation order with a big time.
Here is what really happens:

The expression is evaluated from left-to-right

'k++' increment k (Which is zero) thus becoming 1, but returning 0

'++k' increment k (Which is one) then returning 2

'k++' increment k (which is two) thus becoming 3, but returning 2

The final order of the evaluation is : 0 + 2 + 2

Finally 4 is assigned to k

Not actually what you had in mind, right? Nice try though.

[Davy Kelly]

thank you for correcting me,
so what your saying is, that when the k++ happens, it increments the value k, after the next +, so therefore it increments before the end of the statement!
Davy.
P.s. I will test this out, the wy you look at it, and the way i look at it, to see if i get consistantly same or different output.

[Davy Kelly]

hey Vicken,
cheers, for correcting me, just done a wee program, to see how you looked at it, this is what I done;

public class Foo8 
{
  public static void main (String[] args) 
  {
    int k = 0;
    k = k++ + ++k + ++k + k++ + k++ + ++k + k++;
    System.out.println(k);
  }
}  

7 different increments in the same line, this is how i looked at it in a linear way from left to right.
k = 0
k++ = 0 (but increment to one after)
++k = 2 (used the last increment and the preincrement)
++k = 3 (preincrement)
k++ = 3 (post increment, gets value 3 then adds 1 later)
k++ = 4 (post increment, gets value 3 then adds 1 later)
++k = 6 (used the last increment and the preincrement)
k++ = 6 (post increment, gets value 3 then adds 1 later but does not really get added)
so the value k = 0 + 2 + 3 + 3 + 4 + 6 + 6 which is
24
I think i get it better now, correct me if I am wrong, but this week has been hell, I am soo tired I can't think straight
Davy

[Vicken Karaoghlanian]

Congratulations!!! You hit the jackpot.
To simplify things even more.

Preincrement operator : increment then return value.

Postincrement operator: return value then increment.

Your week being hell!!! You should see mine, I have my exam set next week, I haven't slept for... let me see two weeks maybe. I arrive late to my work place almost every morning, my boss is this close to fire me.
It is really hard to study and work at the same time.

[Davy Kelly]

Vicken,
I read you loud and clear mate, I too have my exam next monday, and I have an interview to prepare for for this wednesday for promotions in the work (nothing to do with IT) and I am working full time 7 days a week. but at least i get a few days off after the 7.
Good luck with your exam, mate
Davy