CODEWITHSUNDEEP
javaarraysvariadic-functions
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Reputation: 13

An array of Strings vs String Varargs

What is the difference between void method(String[] a) and void method(String... a)?

The first method take an array of Strings where as the second method takes one or more String arguments. What are the different features that they provide?

Moreover, don't know why but this is working:

public class Test {

    public static void main(String[] args) {
        String[] a = {"Hello", "World"};
        Test t = new Test();
        t.method(a);
    }

    void method(String...args) {
        System.out.println("Varargs");        // prints Varargs 
    }
}

Upvotes: 1

Views: 6617

Answers (3)

Eritrean
Eritrean

Reputation: 16498

Not nice, but sometimes very usefull: If at time of writing your code you do not know which args your method gets as input your code would look like

public class Test {
public static void main(String[] args) {        
    Test t = new Test();
    String[] a = {"Hello", "World"};
    Integer [] b = {1,2};
    t.method(a);
    t.method(b);
    t.method('a',1,"String",2.3); 
}
void method(Object...args) {
    for (Object arg : args){
    System.out.println(arg);        
    }        
}

Upvotes: 0

Ahmad Al-Sanie
Ahmad Al-Sanie

Reputation: 3785

The only difference is when you invoke the method, and the possible ways to invoke it as varargs allow multiple strings separated by "," or the use of an array as a parameter.

Important Note:

varargs feature automates and hides the process of multiple args passed in an array.

Documentation.

Edit: Java doesn't allow this public void method(String ... args, String user) as varargs must be the last parameter to be called.

Upvotes: 2

Stultuske
Stultuske

Reputation: 9427

There is no difference, only if there are other elements after it in the signature.

For instance:

public void method(String[] args, String user){}

is possible, since there is no way the jvm will think user is still an element of args.

public void method(String ... args, String user){}

will cause trouble, though.

Upvotes: 8

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