Operator doubt

[Asad Zubair]

Hi,

The following code snippet prints 15. I was expecting it to print 16.

int s = 5;
s += s + s++;
System.out.println(s);

It seems like it is not using the post increment (s++) to increment s by 1 after evaluating s as (s=5+5+5 = 15). This is surprising to me as s is being printed in the next line and I thought that would be enough for the post increment to kick in before the printing occurred. Could someone please clarify my doubts here ?

[Ryan Beckett]

The post-increment operator executes after the expression is evaluated. It executes after s is assigned its new value. It can be re-written as follows.

int s = 5;
s += s + s; //s is 15 after this statement executes.
s += 1;
System.out.println(s);

[Chaitanya Kidambi]

Asad Zubair wrote:Hi,

The following code snippet prints 15. I was expecting it to print 16.

int s = 5;
s += s + s++;
System.out.println(s);

It seems like it is not using the post increment (s++) to increment s by 1 after evaluating s as (s=5+5+5 = 15). This is surprising to me as s is being printed in the next line and I thought that would be enough for the post increment to kick in before the printing occurred. Could someone please clarify my doubts here ?

Hello Asad
try to put the whole expression in System.out.println() you will get 15 as a result.

System.out.println(s+=s+s++);

[Asad Zubair]

Guys, I believe you have missed the point in my question. Why does my code print 15 and not 16 ? Should the post increment not kick in after line 2 is finished but before line 3 is executed ?

Ryan, your re-written code is printing 16, not 15. Even though I thought it would execute the same way as my code (which prints 15)

[Matthew Brown]

If you check the Java Language Specification, you'll find the following under "Compound Assignment Operators".

First, the left-hand operand is evaluated to produce a variable...

...the value of the left-hand operand is saved and then the right-hand operand is evaluated...

...the saved value of the left-hand variable and the value of the right-hand operand are used to perform the binary operation indicated by the compound assignment operator

In other words, it uses the value of s at the start of processing (5) to add to the result of the right-hand operation (10) to produce the final result to assign (15). The incremented value is therefore not used in the calculation, and is overwritten by the final assignment.

[Asad Zubair]

Thanks Matthew,
It still does not make total sense to me why the post increment gets overwritten rather than being evaluated after line 2 completes. However, looks it is a JLS rule and I just need to burn it in for the exam.

[Ivan Amat]

Hi,

I think that s += s + s++; is the same as say s = s + (s + s++);

step 1: s = s + (s + s++);
step 2: s = 5 + (5 + 5);
step 3: s = 5 + (10); // in this moment, do s++ and s = 6
step 4: s = 15 // overwritting the value of 6 that has 's'

I hope help you.

Sorry for my english

[Asad Zubair]

As I start to play with the code a bit more, it makes more sense now. The execution happens differently for comparison operators vs assignment operators (due to the over-writing concept). I previously thought it was the same. Anyways, nice little exam gotcha that I should be prepared for now. Thanks for the comments !

int s = 5;
if(s > s++){			//this is a comparison not an assignment so the post increment does not get overwritten
	System.out.println("greater");
}else{
	System.out.println("less than or equal to");	//this gets printed
}
System.out.println(s);	//s has been post-incremented to 6
s = s++; 				//this is an assignment so the post increment gets overwritten
System.out.println(s); 	//s is still 6
s = ++s;
System.out.println(s); 	//s is now 7, the pre-increment gives it a chance to save before getting overwritten by the assignment