CODEWITHSUNDEEP
java
Korben
Korben

Reputation: 736

Java method overloading parameter management

suppose I have a two methods defined in on class as following

public void m(final Serializable s){
    System.out.println("method 1");
}

public void m(final String s){
    System.out.println("method 2");
}

if I have a variable defined as

final Serializable s = "hello world";

How can I use this variable to invoke m so that "method 2" will be printed on the console? I have tried

m(s.getClass().cast(s));

but still method 1 is invoked, but the above code does cast s to String type from my observation.

Upvotes: 1

Views: 88

Answers (3)

BUY
BUY

Reputation: 736

You can cast it to a String in several ways like:

m((String) s);

or:

m(s.toString());

Or you can initialize a String variable with s and invoke the method with it like:

String invoker = "" + s;
m(invoker);

Upvotes: 0

Daniele
Daniele

Reputation: 2837

Your variable s is of type Serializable, but in fact it points to a String object (since the literal "hello world" gives you an object of that type), so it can be cast to (String).

Overloaded method calls are resolved at compile time using the most specific type available (known at compile time); what you are doing here, is trying to resolve using the most specific type known at run-time.

In Java you could do that using the Visitor pattern, (but unfortunately you can't extend String; you can if you declare your own types).

Upvotes: 4

Robert Kock
Robert Kock

Reputation: 6038

Try this:

if (s instanceof String)
  m((String)s);
else
  m(s);

Upvotes: 1

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