Confused

[Arun C. Giridharan]

javachamp Q1

 public class Tester {
     
   static void call(Long x, Long y) {
        System.out.print("Long x, Long y");
    }
 
   static void call(int... x) {
       System.out.print("int... x");
   }

  public static void main(String[] args) {
       int val = 3;
        call(val, val);
    }
}   

1 Long x, Long y
2 int... x
3 Compilation error
4 An exception is thrown at run time




The Answer was static void call(int... x) .....how is that possible? Can anyone Explain me!

[Wouter Oet]

This topic will explain it.

[David Newton]

What's even the question?

[Arun C. Giridharan]

Thank you Wouter Oet.

[Campbell Ritchie]

David Newton wrote:What's even the question?

And please put some hint of the question in the thread title.

[David Newton]

Campbell Ritchie wrote:And please put some hint of the question in the thread title.

Good point.

Technically every single post I've ever made in my life could be titled "Confused."

[Shanky Sohar]

public class Tester {

static void call(Long x, Long y) {
System.out.print("Long x, Long y");
}

static void call(int... x) {
System.out.print("int... x");
}

public static void main(String[] args) {
int val = 3;
call(val, val);
}
}

compiler choose..
1.matching argument
2.widening
3.autoboxing.
4.Variable argument.

if you remove

static void call(int... x) {   
       System.out.print("int... x");   
   }

then compiler says compile time error.on call(val,val);
because it is not legal to first autobox(int to Long) and then widened it to long.

but if you change

static void call(Long x, Long y) {   
        System.out.print("Long x, Long y");   
    }

with this

static void call(long x, long y) {    //converted Long wrapper class to long primitive type 
            System.out.print("long x, long y");   
        }

then output will be "long x, long y"
because it is perfectly legal to widening.



So,In your code
compiler doesnot even see the code other call(Long x,Long y) because it is finding other one to match which is

static void call(int... x) {   
       System.out.print("int... x");   
   }    

that why output is "int....x"

[Vishal Kashyap]

Keep It Up Shanky..........Well Done.........

[Shanky Sohar]

Vishal Kashyap wrote:Keep It Up Shanky..........Well Done.........