println does not print the expected value

Question

This is my code:

public static void main(String[] arg)
{

    String x = null;
    String y = "10";
    String z = "20";

    System.out.println("This my first out put "+x==null?y:z);

    x = "15";

    System.out.println("This my second out put "+x==null?y:z);

}

My output is:

20
20

But I'm expecting this:

This my first out put 10
This my second out put 20

Could someone explain me why I'm getting "20" as output for both println calls?

Funest

Sumit Singh's answer explains why the output is like it is. The reason is operator precedence, see this document – Sugarcoat

Answer 1

System.out.println("This my first out put "+x==null?y:z); will be executed like

("This my first out put "+x)==null?y:z which is never going to be true. So, it will display z value.

For example:

int x=10;
int y=20;
System.out.println(" "+x+y); //display 1020
System.out.println(x+y+" "); //display 30

For above scenario, operation performed left to right.

As, you said, you are expecting this:

This my first output 10

For this, you need little change in your code. Try this

System.out.println("This my first output " + ((x == null) ? y : z));

Jemine

evaluation of expressions in Java is NOT always left to right. It takes account of operator precedence! – Zouave

@StephenC i know, but i was talking for this scenario only. Rather, it will increase more confusion, i removed ALWAYS from my statement. I believe, now it will be OK. :) – Jemine

Answer 2

Try

System.out.println("This my first out put "+ (x==null?y:z));

Tentacle

This will indeed generate the desired output, but does not answer OPs question: Could someone explain me why I'm getting "20" as output for both println calls? – Sugarcoat

Answer 3

you need to try:

System.out.println("This my first out put "+(x==null?y:z));
x = "15";
System.out.println("This my second out put "+(x==null?y:z));

Rattlebrain