Java Object Oriented Design - Java varargs Method

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The term "varargs" is shorthand for "variable-length arguments."

The varargs declares a method or constructor that accepts a variable number of arguments (or parameters).

Declare varargs Method

To declare varargs, add an ellipsis ... after the data type of the method's argument.

The following code shows a max() method declaration with one variable-length argument, num, which is of the int data type.

public static  int max(int... num) {

}

Adding whitespaces before and after ellipsis is optional.

A varargs method can have more than one argument. The following code shows that aMethod() accepts three arguments, one of which is a variable-length argument:

public static int aMethod(String str, double   d1,   int...num)  {

}

A varargs method can have a maximum of one variable-length argument.

The variable-length argument of a varargs method must be the last argument in the argument list.

void  m2(String str, int...n1) {

}

Example

Let's rewrite the max() method to make it a varargs method:

public class Main {
  public static int max(int... num) {
    int max = Integer.MIN_VALUE;
    for (int i = 0; i < num.length; i++) {
      if (num[i] > max) {
        max = num[i];//from  ww  w.j  a  v a  2  s  . com
      }
    }
    return max;
  }
}

Use varargs Method

We can use a for loop to process the list of arguments for the variable-length argument.

The length property gives you the number of values that were passed for the variable-length argument.

To get the nth value in the variable-length argument, you need to use varArgsName[n-1].

We can use a foreach loop to work with variable-length argument.

public class Main {
  public static int max2(int... num) {
    int max = Integer.MIN_VALUE;
    for (int currentNumber : num) {
      if (currentNumber > max) {
        max = currentNumber;
      }
    }
    return max;
  }
}

We can call the Main.max() method as follows:

int max1 = Main.max(1, 8);
int max2 = Main.max(1, 1, 3);

You can use zero or more arguments for a variable-length argument in a method. The following code is a valid call to the max() method:

int max = Main.max(); // Passing no  argument  is ok

The following declaration of the max() method will force its caller to pass at least two integers:

// Argumenets  n1  and  n2  are   mandatory
public static int max(int n1,  int n2,   int... num) {

}

The compiler will treat the first two arguments, n1 and n2, as mandatory and the third argument, num, as optional.

public class Main {
  public static int max(int n1, int n2, int... num) {
    // Initialize max to the maximum of n1 and n2
    int max = (n1 > n2 ? n1 : n2);
/*ww w .  j  av  a2  s. c  o m*/
    for (int i = 0; i < num.length; i++) {
      if (num[i] > max) {
        max = num[i];
      }
    }
    return max;
  }

  public static void main(String[] args) {
    System.out.println(max(7, 9));
    System.out.println(max(7, 9, 10));
    System.out.println(max(7, 9, 10, 13));
  }
}

The code above generates the following result.

Overloading a Varargs Method

The same overloading rules for methods apply to a varargs method.

We can overload a method with a variable-length argument as long as the parameters for the methods differ in type, order, or number.

For example, the following is a valid example of an overloaded max() method:

public  class  Main {
    public static int max(int x, int  y)  {
    }
    
    public static int max(int...num)  {
    }
}

Consider the following code

int max = Main.max(12, 13);  // which  max()  will be  called?

Java will call the max(int x, int y). Java first attempts to find a method declaration using an exact match for the number of parameters. If it does not find an exact match, it looks for a match using variable-length parameters.

If a varargs method is overloaded, Java uses the more specific version of the method instead of using a varargs method. java uses varargs method as the last resort to resolve a method call.

The overloading of the method itself may be valid. However, the call to it may cause an issue.

public class Main {
  public static int max(int... num) {
    return 0;
  }

  public static int max(double... num) {
    return 0;
  }
}

Which max() would the following code call?

int max = Main.max(); // Which max()  to call?

The above statement will generate a compilation time error.

Varargs Methods and the main() Method

The signature for the main() method must be main(String[] args).

The following declaration of main() method for the Main class is valid.

public class Main {
  public static void main(String... args) {
    System.out.println("Hello from  varargs main()...");
  }
}

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