Java Default Values

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Question : Default parameters in Java

Answered by : beautiful-bug-r6w4y2smd5wt

// 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!
*/

Source : | Last Update : Sat, 26 Feb 22

Question : java default values

Answered by : santosh-pal

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.

Source : | Last Update : Thu, 18 Mar 21

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