Why is the output of the following code...?

Question

public class Yikes {
 public static void go(Long n) {System.out.println("Long ");}
 public static void go(Short n) {System.out.println("Short ");}
 public static void go(int n) {System.out.println("int ");}
 public static void main(String [] args) {
 short y = 6;
 long z = 7;
 go(y);
 go(z);
 }
}

the output of the above code is int long but why and how?

Jimmy S · Accepted Answer

No, because it's Java and the S on Short is capitalized.
Try this code out with your variable types in lowercase.

public class Yikes {
	public static void go(long n) {System.out.println("Long ");}
	public static void go(short n) {System.out.println("Short ");}
	public static void go(int n) {System.out.println("int ");}
	public static void main(String [] args) {
	short y = 6;
	long z = 7;
	go(y);
	go(z);
	}

}

Gardner · Answer

Because the values 6 or 7 can be long or short.  It is running the first GO that it finds and it reports that the number is long.

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Why is the output of the following code...?

2 Answers