Vararg Problem

[Mohit G Gupta]

public class TestMain { 

    public static void main(String[] args) {
        printName("Skill","Test");

    }

 

    public static void printName(String firstName,String lastName){

        System.out.println("In Method Id 100");

    }

 

     public static void printName(String... name){

        System.out.println("In Method Id 900");

    }

}
    

this code runs fine


with output:
In Method Id 100 will be printed

i think output should be
Compile time error as it is not clear which method can be called


can anyone tell the reason for this ?

[Dieter Quickfend]

var-arg has the lowest precedence. If a program has a choice, it will pretty much always choose anything else over a var-arg. The var-arg is the last resort, because it's least specific. It's like passing a Dog where a method is overloaded to accept either Object or Animal. Object is least specific, so it will choose Animal. Var-args behave the same way. They are the least specific, because if you want, you can pass 100 arguments.

Check out K&B study guide, it clearly states precedence rules. It's in Chapter 3, I believe. Not sure though. But it's in there.

[Abimaran Kugathasan]

mohitkumar gupta wrote:

i think output should be
Compile time error as it is not clear which method can be called

Why did you think so?

[Arjun Srivastava]

why it should be the compiler error?
perhaps you are thinking static method can't be override but here method is overloaded.
remember varag is a loser in each and every case

[Wouter Oet]

This topic is a must read about auto-boxing, boxing and var-args.

[Shanky Sohar]

above topic will be helpful.

but just remember one thing
compiler choose
widening,
autoboxing and then
var-args.

[Sven Mathijssen]

arjun srivastava wrote:
why it should be the compiler error?
perhaps you are thinking static method can't be override but here method is overloaded.
remember varag is a loser in each and every case

I understand the confusion, because the call to printName(String, String) is ambiguous in this case. However, the post mentioned by Wouter Oet clearly states the compiler resolving this ambiguity by choosing the declared two-argument method over the varargs variant. In this case it is resolved by precedence of a fixed-number-of-arguments method over a method with varargs.

[Akshat Kumar]

From what i know,
String,String seems like a perfect match.

I think the var args method will be chosen if you use an invocation with just 1 String or more than 2 Strings.

[Shanky Sohar]

Welcome to JavaRanch.

For better understanding have a look at above posts.

[Gari Jain]

Var-args is the last preference.
(K&B)

read the var-args topic