177

Is it possible to declare a method that will allow a variable number of parameters ?

What is the symbolism used in the definition that indicate that the method should allow a variable number of parameters?

Answer: varargs

Pavle Šarenac
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user69514
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    Since its homework, we don't want to know your question, we just want to know you are learning. – Dave Nov 18 '13 at 17:19

6 Answers6

313

That's correct. You can find more about it in the Oracle guide on varargs.

Here's an example:

void foo(String... args) {
    for (String arg : args) {
        System.out.println(arg);
    }
}

which can be called as

foo("foo"); // Single arg.
foo("foo", "bar"); // Multiple args.
foo("foo", "bar", "lol"); // Don't matter how many!
foo(new String[] { "foo", "bar" }); // Arrays are also accepted.
foo(); // And even no args.
ErikE
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BalusC
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    Is it possible to do various type of paramteres? e.g. (String...strs, int... ints). What about just any type of argument in any order? – trusktr Oct 04 '13 at 01:49
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    @trusktr: if you want any object, just use `Object...`. – BalusC Oct 04 '13 at 01:51
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    @trusktr No, primitives are not objects. There is a great explanation of the difference here: http://www.programmerinterview.com/index.php/java-questions/difference-between-a-primitive-type-and-a-class-type/ – Dick Lucas Aug 03 '14 at 15:25
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    @Richard: Using `Object... args` will work with primitives because of [autoboxing](https://docs.oracle.com/javase/tutorial/java/data/autoboxing.html). – Sumit May 02 '16 at 15:37
  • How would you check the number of arguments? – Stevoisiak Apr 30 '17 at 18:11
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    @StevenVascellaro : I assume that the arguments are handled like an array, so you could probably simply do `varargs.length` – Luatic May 25 '17 at 14:23
18

Yes, it's possible:

public void myMethod(int... numbers) { /* your code */ }
Skaldebane
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Dolph
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17
Variable number of arguments

It is possible to pass a variable number of arguments to a method. However, there are some restrictions:

  • The variable number of parameters must all be the same type
  • They are treated as an array within the method
  • They must be the last parameter of the method

To understand these restrictions, consider the method, in the following code snippet, used to return the largest integer in a list of integers:

private static int largest(int... numbers) {
     int currentLargest = numbers[0];
     for (int number : numbers) {
        if (number > currentLargest) {
            currentLargest = number;
        }
     }
     return currentLargest;
}

source Oracle Certified Associate Java SE 7 Programmer Study Guide 2012

Victor Sergienko
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Abey Ella
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14

For different types of arguments, there is 3-dots :

public void foo(Object... x) {
    String myVar1  = x.length > 0 ? (String)x[0]  : "Hello";
    int myVar2     = x.length > 1 ? Integer.parseInt((String) x[1]) : 888;
}

Then call it

foo("Hii"); 
foo("Hii", 146);

for security, use like this:
if (!(x[0] instanceof String)) { throw new IllegalArgumentException("..."); }

The main drawback of this approach is that if optional parameters are of different types you lose static type checking. Please, see more variations .

T.Todua
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6

Yup...since Java 5: http://java.sun.com/j2se/1.5.0/docs/guide/language/varargs.html

Roland Bouman
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0

Yes Java allows vargs in method parameter . 

public class  Varargs
{
   public int add(int... numbers)
   { 
      int result = 1; 
      for(int number: numbers)
      {
         result= result+number;  
      }  return result; 
   }
}
Sasikumar Murugesan
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Code_Eat_Sleep
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