Java variable number or arguments for a method

Question

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

Answer 1

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.

Answer 2

Yes, it's possible:

public void myMethod(int...numbers) { ... }

Answer 3

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

Answer 4

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

public void foo(Object... x) {
    String first    =  x.length > 0 ? (String)x[0]  : "Hello";
    int duration    =  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 .

Answer 5

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

Answer 6

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; 
   }
}