operator presedence

[Minhaj Mehmood]

public class Test{
	public static void main(String[] args) {
		int i = 5;
		doCal(i);
	}
	public static final int AN_INT = 3;
	public static void doCal(int x) {
		++x;
		System.out.println(x += AN_INT + ++x); //why it`s printing 16 here  :confused: 
	}
}

I found it in jqplus.

[rohan yadav]

Expression is evaluated to x = (x) +(AN_INT+(++x))
that's why output is 16

[Minhaj Mehmood]

the lass ++x wont effect the actual value of x??

[Ankur kothari]

i think its because of ( x += AN_INT)+((++x))) becasue +comes first in the hierarchy

[Ankur kothari]

  int x=7;
x*=7+5;
wot is x now?  

here x is 84....according to K&B right of += is calculated first...

so minhaj i think your answer should be x+=3+7; which is x=7+10...17

then why 16? hmmmm

[Chandana Garlapati]

Minhaj kaimkhani wrote:

public class Test{
	public static void main(String[] args) {
		int i = 5;
		doCal(i);
	}
	public static final int AN_INT = 3;
	public static void doCal(int x) {
		++x; // here x becomes 6
		System.out.println(x += AN_INT + ++x); // here x=(x) + (AN_INT) + (++x) = 6 + 3 + 7 =16
	}
}

Hope its clear now..

[Ankur kothari]

then why is x*=7+5; not x=x*7+5; ?

[Ankur kothari]

maybe its becasue its x*(7+5) becasue of that rule agian

[Minhaj Mehmood]

Chandana Garlapati wrote:

Minhaj kaimkhani wrote:

public class Test{
	public static void main(String[] args) {
		int i = 5;
		doCal(i);
	}
	public static final int AN_INT = 3;
	public static void doCal(int x) {
		++x; // here x becomes 6
		System.out.println(x += AN_INT + ++x); // here x=(x) /*<= why this x is not incremented  because of last ++x*/ + (AN_INT) + (++x) = 6 + 3 + 7 =16
	}
}

Hope its clear now..

[Ankur kothari]

maybe its becasue the operands are evaluated from left to right....

i think this is x+= (AN_INT) + (++x).....first += is opened up soit is

its x=(x) + (AN_INT) + (++x)

this is simply so confusing

[Chandana Garlapati]

Assignment operators have right to left associativity.i.e; the expression is evaluated from right to left.

[Ankur kothari]

so the output should be 17 right Chandana?

[Neha Daga]

No, the answer is 16.

when post increment is applied second time the value is stored temporarily and after all calculations final value is assigned to x.

[Ankur kothari]

neha can you explain in detail?

[Chandana Garlapati]

ones at the right should be performed before the ones at the left.

[Ankur kothari]

ok so correct me if am wrong

here x is 6;
x+=3 + ++x;

so Chandana according to you it would be x+=3+ 7; which evaluates to x+=10;

which is x=x+10;

which is x=10+7; right? or 10+6?

[Chandana Garlapati]

As Neha said, value is stored temporarily and after performing the right expression then x value is added...

lets try this

  

System.out.println(x = AN_INT + ++x+ x);
  

Now the output is 17..

[Ankur kothari]

leave it...why dont either of you simply show me the calculation step by step...the answer is 16....maybe Ankit or Henry could explain properly

[Neha Daga]

ok..expression is x+= AN_INT + ++x which is x = x + AN_INT + ++x
now according to the rule right side will be calculated first but from left to right
so it becomes: 6+ 3 +7=16.

now, I hope its clear.

[Raju Champaklal]

i just read the jls and there it is mentioned that the program rememebrs what you had on the left hand side...so its like x+=3+7 whcih gives x+=10 which is x=x+10 equals x=6+10 which is 16......Bingo

understood Ankur???

[Raju Champaklal]

  public class And 
{



public static void main(String[] args) {
		int i = 2;int j=3;
		 j *=j++ + ++j; 
		System.out.println(j);// j is 24..j *=8; which gives j=3*8 becasue j is 3 here..the program remebered the value of j :thumbup: 
	}

}