Operators

[zohaib khan]

hi guys,

Following code produces result as 11.

I have confusion about it.

Can anybody give me short explanation about it.

I think output should be 12.
thanks in advance
class arithmaticTest
{
public static void main(String []args)
{
int i=10;
i=++i;
i=i++;
System.out.println(i);
}
}

[Ankit Garg]

Zohaib, when you post a source code, please Use Code Tags so that the code is readable.

class arithmaticTest
{
    public static void main(String []args)
    {
        int i=10;
        i=++i;  //i is 11 now
        i=i++;  //problematic line
        System.out.println(i);
    }
} 

The question that you've asked is simple. I've commented the line where i will be 11 in the code. The next line is a bit confusing at first. The steps that will happen is this

i = i++;

first the part in the bold will be evaluated. Since this is a postfix increment operator, so the expression will result in 11. By expression I mean it will be equivalent to

i = 11;

But the assignment will not be made first. First the value will be increased. Then i will be assigned 11. So you can think of that line as

i = 11;
temp = i;
i++;
i = temp;

I hope it makes sense to you...

[Seetharaman Venkatasamy]

Please use code tag to post you code to make clear. Coming to your question

try(run) this and compare to your code

    
 class arithmaticTest
{
public static void main(String []args)
{
int i=10;
i=++i;
i++; // in your code i = i++; it assign 11 to i since it is post increment
System.out.println(i);
}
} 

[Seetharaman Venkatasamy]

Ohh Ankit beat me here . i dont see your post while i post

[Abhay Agarwal]

The reason it behaves as observed is because of the interaction between the operator precedence and the odd "return value" of the ++ operator.
The postfix increment happens first (see your favorite copy of the operator precedence chart); but postfix, by definition, returns the previous value of the variable. So that "return value" is not "original i plus one", but instead just "original i" (i, at this point, actually contains the incremented value). This returned value is then assigned to i, which overwrites what the ++ did in the first place.

[Abhay Agarwal]

Well you have beaten me also ... seems that i need to increase my typing speed...

[Ankit Garg]

seetharaman venkatasamy wrote:Ohh Ankit beat me here . i dont see your post while i post

Abhay Agarwal wrote:Well you have beaten me also ... seems that i need to increase my typing speed...

Yepee, I came first

[Bob Wheeler]

Please put the question with the code brackets.
If you look at the following line

 i=i++;   

What happens? We have here a postfix increment operator. But what is the precedence between increment operator and '=' operator? The '=' assignment operator has the least precedence !!!
So what happens into more detail: First we copy the value of i (here 11) to memory, then we increment i (to 12) and as the last operation we assign the copied value to i, so i is back to 11.
So you see, if we assign the incremented value of i to another variable, let say j, then the output would be 12, like here:

 
public class arithmaticTest
{
	public static void main(String []args)
	{
		int i=10,j = 0;
		i=++i;
		j=i++;
		System.out.println(i);
	}
} 
   

At least, that's how I see the pre/postfix operators.
Hope this helps.
cheers Bob