+ 4
I got somethin funny. Here it is
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); } } when run the output is- 6 5 my problem is if execution starts from top to bottom why is not the output 5 6. help me
9 Answers
+ 18
statement(s) in main method are executing from top to bottom only , see carefully
//thats why "6\n5\n" as output
+ 8
hilarious! are you a comedian?
+ 3
gotcha tank you
+ 2
It seems to be because of the order of output.
The System.out.println after calling addOneTo is printing out the value of x inside main().
The System.out.println call in addOneTo is also printing it's local value of x, but after adding 1 to it.
So, if you swap those two lines you should be fine.
+ 2
Ha! you're right, that is funny!
why the downvote!
+ 2
The execution starts in main method, the first call is to addOneTo(x) this one recives x as a parameter and print x+1. The next instruction called is System.out.println(x), this one print x.
+ 1
De todos modos me parece que la idea del ejercicio era mostrar el 'ambito de trabajo' de la variable, ya que aunque el metodo addOneTo(x) incrementa el valor de su parametro, la variable x del metodo main no se altera.
+ 1
Bécasse the statement in main() executes first and second statement exécutés after vallon.
0
if you sumation num5 & num1 in binary
for exampl:
1---> 0001
+
5--->0101
= 0101=5