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?

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));

Answer 2

尝试

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

Answer 3

use following code this will solve your problem: The probelm is because its taking -

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

As

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

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));

}

Answer 4

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));