JAVA

default argument in java

Short answer: No.

Fortunately, you can simulate them.
  
-  use method overloading
-  allow nulls as an input
-  declare a method with Java Varargs

To know how to do those, check : 
	http://dolszewski.com/java/java-default-parameters/

Default parameters in Java

// Foo.java
public class Foo {
    // For a constructor
    public Foo() { this(4); }
    public Foo(int x) {
        System.out.println("Initializing Foo with " + x + ".");
    }

    // For a method
    public void identifyString() { identifyString("Grepper is cool!"); }
    public void identifyString(String source) {
        int length = source.length();

        if (length < 0)
            System.out.println("what the heck");
        else if (length == 0)
            System.out.println("Source is empty!");
        else if (length == 1)
            System.out.println("Source is a char!");
        else
            System.out.println("Source is a String!");
    }
}

// Main.java
public class Mainf {
    public static void main(String[] args) {
        Foo foo = new Foo();
        foo.identifyString();
    }
}


/* OUTPUT:
    Initializing Foo with 4.
    Source is a string!
*/

java default values

No, the structure you found is how Java handles it (that is, with overloading instead of default parameters).

For constructors, See Effective Java: Programming Language Guide's Item 1 tip (Consider static factory methods instead of constructors) if the overloading is getting complicated. For other methods, renaming some cases or using a parameter object can help. This is when you have enough complexity that differentiating is difficult. A definite case is where you have to differentiate using the order of parameters, not just number and type.