is this strange

[kashif sohail]

(code/)
public class Qcb90 {
int a;
int b;
public void f() {
a = 0;
b = 0;
int[] c = { 0 };
g(b, c);
System.out.println(a + " " + b + " " + c[0] + " ");
}
public void g(int b, int[] c) {
a = 1;
b = 1;
c[0] = 1;
}
public static void main(String args[]) {
Qcb90 obj = new Qcb90();
obj.f();
}
}
(code/)
as i know primitive type are reference by copy so what wil be the result and why
any help

[Vlad G]

Output: 1 0 1
Why:
in g()
a = 1; - set class member variable a to 1
b = 1; - set local variable b to 1, not the member var. b which stays 0 (this.b = 1 would set member var b)
c[0] = 1; - arrays are objects and passed by reference, so this sets first element of array created in f() to 1.
-VG

[kashif sohail]

actually what confusing me is a an b are primtives and a copy of primitives is given to each caller then how value is 101
as each method has its own copy
bit confused
plz help
kashif

[Cindy Glass]

There is only one place in your example that you pass anything. That is in calling the method g.
You only passed b and c, and c is an object not a primitive. That is EXACTLY the reason that you get different results when printing a than when printing b.
When method g worked on a it worked on the object variable and set it equal to 1.
When it worked on b it worked on the local variable (copy) so the print statement prints the object variable which is still 0.