Java method overloading parameter management

Question

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.

m((String)s); or m(s.toString()) if you want this to work for all reference types. — Eran
– Eran, Commented Jun 6, 2018 at 7:52
instead of final Serializable s = "hello world"; use final String s = "hello world"; ot cast the param — Dmitri Gudkov
– Dmitri Gudkov, Commented Jun 6, 2018 at 7:53
@Eran I can't do explicit casting since there are other variables like Integer.. etc — Korben
– Korben, Commented Jun 6, 2018 at 7:53
Possible duplicate of Overloaded method selection based on the parameter's real type — Ryan Leach
– Ryan Leach, Commented Jun 6, 2018 at 7:54

Daniele · Accepted Answer · 2018-06-06 08:02:07Z

4

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).

answered Jun 6, 2018 at 8:02

Daniele

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1 Comment

BUY Over a year ago

A nice and clean answer, clarifying the problem. Good job.

Robert Kock · Accepted Answer · 2018-06-06 07:56:25Z

1

Try this:

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

answered Jun 6, 2018 at 7:56

Robert Kock

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Comments

BUY · Accepted Answer · 2018-06-06 08:05:22Z

0

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);

answered Jun 6, 2018 at 8:05

BUY

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