WHY output is 6, 5 instead of 5,6. How the value 5 assigned to num? For below code.

Question

public class MyClass {
public static void main(String[ ] args) {
int x = 5;
addOneTo(x);
System.out.println(x);
}

static void addOneTo(int num) {
num = num + 1;
System.out.println(num);
}
}

Answer 1

Important: void addOneTo() does not return a value. The method gets a value, increment it and print it. This has no effect to your variable x.

x = 5
then you call addOneTo(5)
num = 5 + 1 = 6
print num -> print 6
After that you jump back to your main method:
print x -> x is still 5, so print 5

Output:
6
5

If you want a method which changes x:

public static int addOneTo(int num){

return num + 1;

}

inside main:

int x = 5;
x = addOneTo(x);
System.out.println(x);

Output: 6

Answer 2

The code will display the content in the order its compiled so

int x = 5; //x equals 5
addOneTo(x)// grab a copy of the value of x above me and send it over to that method outside the main method, add one to it and print 6 to console once your done come back and finish executing the code in the main method
System.out.println(x)// since only a copy was sent to the addOneto method then x still remains being the value of 5 so 5 is printed to console.

6
5 is printed to console

Denise Roßberg explained this very well already