Method overloading

Bruce Eckel’s Thinking in Java Contents | Prev | Next

One
of the important features in any programming language is the use of names. When
you create an object, you give a name to a region of storage. A method is a
name for an action. By using names to describe your system, you create a
program that is easier for people to understand and change. It’s a lot
like writing prose – the goal is to communicate with your readers.

You
refer to all objects and methods by using names. Well-chosen names make it
easier for you and others to understand your code.

A
problem arises when mapping the concept of nuance in human language onto a
programming language. Often, the same word expresses a number of different
meanings – it’s
overloaded.
This is useful, especially when it comes to trivial differences. You say
“wash the shirt,” “wash the car,” and “wash the
dog.” It would be silly to be forced to say, “shirtWash the
shirt,” “carWash the car,” and “dogWash the dog”
just so the listener doesn’t need to make any distinction about the
action performed. Most human languages are redundant, so even if you miss a few
words, you can still determine the meaning. We don’t need unique
identifiers – we can deduce meaning from context.

Most
programming languages (C in particular) require you to have a unique identifier
for each function. So you could not have one function called
print( )
for printing integers and another called
print( )
for printing floats – each function requires a unique name.

In
Java, another factor forces the overloading of method names: the constructor
.
Because the constructor’s name is predetermined by the name of the class,
there can be only one constructor name. But what if you want to create an
object in more than one way? For example, suppose you build a class that can
initialize itself in a standard way and by reading information from a file. You
need two constructors, one that takes no arguments (the
default
constructor), and one that takes a
String
as an argument, which is the name of the file from which to initialize the
object. Both are constructors, so they must have the same name – the name
of the class. Thus
method
overloading

is essential to allow the same method name to be used with different argument
types. And although method overloading is a must for constructors, it’s a
general convenience and can be used with any method.

Here’s
an example that shows both overloaded constructors and overloaded ordinary
methods:

//: Overloading.java
// Demonstration of both constructor
// and ordinary method overloading.
import java.util.*;
 
class Tree {
  int height;
  Tree() {
    prt("Planting a seedling");
    height = 0;
  }
  Tree(int i) {
    prt("Creating new Tree that is "
        + i + " feet tall");
    height = i;
  }
  void info() {
    prt("Tree is " + height
        + " feet tall");
  }
  void info(String s) {
    prt(s + ": Tree is "
        + height + " feet tall");
  }
  static void prt(String s) {
    System.out.println(s);
  }
}
 
public class Overloading {
  public static void main(String[] args) {
    for(int i = 0; i < 5; i++) {
      Tree t = new Tree(i);
      t.info();
      t.info("overloaded method");
    }
    // Overloaded constructor:
    new Tree();
  }
} ///:~ 

A
Tree
object can be created either as a seedling, with no argument, or as a plant
grown in a nursery, with an existing height. To support this, there are two
constructors, one that takes no arguments (we call constructors that take no
arguments default
constructors
[17])
and one that takes the existing height.

Distinguishing
overloaded methods

If
the methods have the same name, how can Java know which method you mean?
There’s a simple rule: Each overloaded method must take a unique list of
argument types.

If
you think about this for a second, it makes sense: how else could a programmer
tell the difference between two methods that have the same name, other than by
the types of their arguments?

Even
differences in the ordering of arguments is sufficient to distinguish two
methods: (Although you don’t normally want to take this approach, as it
produces difficult-to-maintain code.)

//: OverloadingOrder.java
// Overloading based on the order of
// the arguments.
 
public class OverloadingOrder {
  static void print(String s, int i) {
    System.out.println(
      "String: " + s +
      ", int: " + i);
  }
  static void print(int i, String s) {
    System.out.println(
      "int: " + i +
      ", String: " + s);
  }
  public static void main(String[] args) {
    print("String first", 11);
    print(99, "Int first");
  }
} ///:~ 

The
two
print( )
methods have identical arguments, but the order is different, and that’s
what makes them distinct.

Overloading
with primitives

Primitives
can be automatically promoted from a smaller type to a larger one and this can
be slightly confusing in combination with overloading. The following example
demonstrates what happens when a primitive is handed to an overloaded method:

//: PrimitiveOverloading.java
// Promotion of primitives and overloading
 
public class PrimitiveOverloading {
  // boolean can't be automatically converted
  static void prt(String s) {
    System.out.println(s);
  }
 
  void f1(char x) { prt("f1(char)"); }
  void f1(byte x) { prt("f1(byte)"); }
  void f1(short x) { prt("f1(short)"); }
  void f1(int x) { prt("f1(int)"); }
  void f1(long x) { prt("f1(long)"); }
  void f1(float x) { prt("f1(float)"); }
  void f1(double x) { prt("f1(double)"); }
 
  void f2(byte x) { prt("f2(byte)"); }
  void f2(short x) { prt("f2(short)"); }
  void f2(int x) { prt("f2(int)"); }
  void f2(long x) { prt("f2(long)"); }
  void f2(float x) { prt("f2(float)"); }
  void f2(double x) { prt("f2(double)"); }
 
  void f3(short x) { prt("f3(short)"); }
  void f3(int x) { prt("f3(int)"); }
  void f3(long x) { prt("f3(long)"); }
  void f3(float x) { prt("f3(float)"); }
  void f3(double x) { prt("f3(double)"); }
 
  void f4(int x) { prt("f4(int)"); }
  void f4(long x) { prt("f4(long)"); }
  void f4(float x) { prt("f4(float)"); }
  void f4(double x) { prt("f4(double)"); }
 
  void f5(long x) { prt("f5(long)"); }
  void f5(float x) { prt("f5(float)"); }
  void f5(double x) { prt("f5(double)"); }
 
  void f6(float x) { prt("f6(float)"); }
  void f6(double x) { prt("f6(double)"); }
 
  void f7(double x) { prt("f7(double)"); }
 
  void testConstVal() {
    prt("Testing with 5");
    f1(5);f2(5);f3(5);f4(5);f5(5);f6(5);f7(5);
  }
  void testChar() {
    char x = 'x';
    prt("char argument:");
    f1(x);f2(x);f3(x);f4(x);f5(x);f6(x);f7(x);
  }
  void testByte() {
    byte x = 0;
    prt("byte argument:");
    f1(x);f2(x);f3(x);f4(x);f5(x);f6(x);f7(x);
  }
  void testShort() {
    short x = 0;
    prt("short argument:");
    f1(x);f2(x);f3(x);f4(x);f5(x);f6(x);f7(x);
  }
  void testInt() {
    int x = 0;
    prt("int argument:");
    f1(x);f2(x);f3(x);f4(x);f5(x);f6(x);f7(x);
  }
  void testLong() {
    long x = 0;
    prt("long argument:");
    f1(x);f2(x);f3(x);f4(x);f5(x);f6(x);f7(x);
  }
  void testFloat() {
    float x = 0;
    prt("float argument:");
    f1(x);f2(x);f3(x);f4(x);f5(x);f6(x);f7(x);
  }
  void testDouble() {
    double x = 0;
    prt("double argument:");
    f1(x);f2(x);f3(x);f4(x);f5(x);f6(x);f7(x);
  }
  public static void main(String[] args) {
    PrimitiveOverloading p =
      new PrimitiveOverloading();
    p.testConstVal();
    p.testChar();
    p.testByte();
    p.testShort();
    p.testInt();
    p.testLong();
    p.testFloat();
    p.testDouble();
  }
} ///:~ 

If
you view the output of this program, you’ll see that the constant value 5
is treated as an
int,
so if an overloaded method is available that takes an
int
it is used. In all other cases, if you have a data type that is smaller than
the argument in the method, that data type is promoted.
char
produces a slightly different effect, since if it doesn’t find an exact
char
match, it is promoted to
int.

What
happens if your argument is
bigger
than the argument expected by the overloaded method? A modification of the
above program gives the answer:

//: Demotion.java
// Demotion of primitives and overloading
 
public class Demotion {
  static void prt(String s) {
    System.out.println(s);
  }
 
  void f1(char x) { prt("f1(char)"); }
  void f1(byte x) { prt("f1(byte)"); }
  void f1(short x) { prt("f1(short)"); }
  void f1(int x) { prt("f1(int)"); }
  void f1(long x) { prt("f1(long)"); }
  void f1(float x) { prt("f1(float)"); }
  void f1(double x) { prt("f1(double)"); }
 
  void f2(char x) { prt("f2(char)"); }
  void f2(byte x) { prt("f2(byte)"); }
  void f2(short x) { prt("f2(short)"); }
  void f2(int x) { prt("f2(int)"); }
  void f2(long x) { prt("f2(long)"); }
  void f2(float x) { prt("f2(float)"); }
 
  void f3(char x) { prt("f3(char)"); }
  void f3(byte x) { prt("f3(byte)"); }
  void f3(short x) { prt("f3(short)"); }
  void f3(int x) { prt("f3(int)"); }
  void f3(long x) { prt("f3(long)"); }
 
  void f4(char x) { prt("f4(char)"); }
  void f4(byte x) { prt("f4(byte)"); }
  void f4(short x) { prt("f4(short)"); }
  void f4(int x) { prt("f4(int)"); }
 
  void f5(char x) { prt("f5(char)"); }
  void f5(byte x) { prt("f5(byte)"); }
  void f5(short x) { prt("f5(short)"); }
 
  void f6(char x) { prt("f6(char)"); }
  void f6(byte x) { prt("f6(byte)"); }
 
  void f7(char x) { prt("f7(char)"); }
 
  void testDouble() {
    double x = 0;
    prt("double argument:");
    f1(x);f2((float)x);f3((long)x);f4((int)x);
    f5((short)x);f6((byte)x);f7((char)x);
  }
  public static void main(String[] args) {
    Demotion p = new Demotion();
    p.testDouble();
  }
} ///:~ 

Here,
the methods take narrower primitive values. If your argument is wider then you
must cast
to the necessary type using the type name in parentheses. If you don’t do
this, the compiler will issue an error message.

Overloading
on return values

It
is common to wonder “Why only class names and method argument lists? Why
not distinguish between methods based on their return values?” For
example, these two methods, which have the same name and arguments, are easily
distinguished from each other:

void f() {}
int f() {}
f();

Default
constructors

//: DefaultConstructor.java
 
class Bird {
  int i;
}
 
public class DefaultConstructor {
  public static void main(String[] args) {
    Bird nc = new Bird(); // default!
  }
} ///:~ 

The
line

new
Bird();

creates
a new object and calls the default constructor, even though one was not
explicitly defined. Without it we would have no method to call to build our
object. However, if you define any constructors (with or without arguments),
the compiler will
not
synthesize one for you:

class Bush {
  Bush(int i) {}
  Bush(double d) {}
}

Now
if you say:

new
Bush();

the
compiler will complain that it cannot find a constructor that matches.
It’s as if when you don’t put in any constructors, the compiler
says “You are bound to need
some
constructor, so let me make one for you.” But if you write a constructor,
the compiler says “You’ve written a constructor so you know what
you’re doing; if you didn’t put in a default it’s because you
meant to leave it out.”

The
this
keyword

If
you have two objects of the same type called
a
and
b,
you might wonder how it is that you can call a method
f( )
for both those objects:

class Banana { void f(int i) { /* ... */ } }
Banana a = new Banana(), b = new Banana();
a.f(1);
b.f(2);

If
there’s only one method called
f( ),
how can that method know whether it’s being called for the object
a
or
b?

To
allow you to write the code in a convenient object-oriented syntax in which you
“send a message to an object,” the compiler does some undercover
work for you. There’s a secret first argument passed to the method
f( ),
and that argument is the handle to the object that’s being manipulated.
So the two method calls above become something like:

Banana.f(a,1);
Banana.f(b,2);

This
is internal and you can’t write these expressions and get the compiler to
accept them, but it gives you an idea of what’s happening.

Suppose
you’re inside a method and you’d like to get the handle to the
current object. Since that handle is passed
secretly
by the compiler, there’s no identifier for it. However, for this purpose
there’s a keyword:
this.
The
this
keyword – which can be used only inside a method – produces the
handle to the object the method has been called for. You can treat this handle
just like any other object handle. Keep in mind that if you’re calling a
method of your class from within another method of your class, you don’t
need to use
this;
you simply call the method. The current
this
handle is automatically used for the other method. Thus you can say:

class Apricot {
  void pick() { /* ... */ }
  void pit() { pick(); /* ... */ }
}

Inside
pit( ),
you
could
say
this.pick( )
but there’s no need to. The compiler does it for you automatically. The
this
keyword is used only for those special cases in which you need to explicitly
use the handle to the current object. For example, it’s often used in
return
statements when you want to return the handle to the current object:

//: Leaf.java
// Simple use of the "this" keyword
 
public class Leaf {
  private int i = 0;
  Leaf increment() {
    i++;
    return this;
  }
  void print() {
    System.out.println("i = " + i);
  }
  public static void main(String[] args) {
    Leaf x = new Leaf();
    x.increment().increment().increment().print();
  }
} ///:~ 

Because
increment( )
returns the handle to the current object via the
this
keyword, multiple operations can easily be performed on the same object.


Calling
constructors from constructors

When
you write several constructors for a class, there are times when you’d
like to call one constructor from another to avoid duplicating code. You can do
this using the
this
keyword.

Normally,
when you say
this,
it is in the sense of “this object” or “the current
object,” and by itself it produces the handle to the current object. In a
constructor, the
this
keyword takes on a different meaning when you give it an argument list: it
makes an explicit call to the constructor that matches that argument list. Thus
you have a straightforward way to call other constructors:

//: Flower.java
// Calling constructors with "this"
 
public class Flower {
  private int petalCount = 0;
  private String s = new String("null");
  Flower(int petals) {
    petalCount = petals;
    System.out.println(
      "Constructor w/ int arg only, petalCount= "
      + petalCount);
  }
  Flower(String ss) {
    System.out.println(
      "Constructor w/ String arg only, s=" + ss);
    s = ss;
  }
  Flower(String s, int petals) {
    this(petals);
//!    this(s); // Can't call two!
    this.s = s; // Another use of "this"
    System.out.println("String & int args");
  }
  Flower() {
    this("hi", 47);
    System.out.println(
      "default constructor (no args)");
  }
  void print() {
//!    this(11); // Not inside non-constructor!
    System.out.println(
      "petalCount = " + petalCount + " s = "+ s);
  }
  public static void main(String[] args) {
    Flower x = new Flower();
    x.print();
  }
} ///:~ 

The
constructor
Flower(String
s, int petals)

shows that, while you can call one constructor using
this,
you cannot call two. In addition, the constructor call must be the first thing
you do or you’ll get a compiler error message.

This
example also shows another way you’ll see
this
used. Since the name of the argument
s
and
the name of the member data
s
are the same, there’s an ambiguity. You can resolve it by saying
this.s
to refer to the member data. You’ll often see this form used in Java
code, and it’s used in numerous places in this book.

In
print( )
you can see that the compiler won’t let you call a constructor from
inside any method other than a constructor.


The
meaning of static

With
the
this
keyword
in mind, you can more fully understand what it means to make a
method
static.
It means that there is no
this
for that particular method. You cannot call non-
static
methods from inside
static
methods
[18]
(although the reverse is possible), and you can call a
static
method for the class itself, without any object. In fact, that’s
primarily what a
static
method is for. It’s as if you’re creating the equivalent of a
global function (from C). Except global functions are not permitted in Java,
and putting the
static
method inside a class allows it access to other
static
methods
and to
static
fields.

Some
people argue that
static
methods are not object-oriented since they do have the semantics of a global
function; with a
static
method you don’t send a message to an object, since there’s no
this.
This is probably a fair argument, and if you find yourself using a
lot
of static methods you should probably rethink your strategy. However,
statics
are pragmatic and there are times when you genuinely need them, so whether or
not they are “proper OOP” should be left to the theoreticians.
Indeed, even Smalltalk
has the equivalent in its “class methods.”


[17]
In some of the Java literature from Sun they instead refer to these with the
clumsy but descriptive name “no-arg constructors.” The term
“default constructor” has been in use for many years and so I will
use that.

[18]
The one case in which this is possible occurs if you pass a handle to an object
into the
static
method. Then, via the handle (which is now effectively
this),
you can call non-
static
methods and access non-
static
fields. But typically if you want to do something like this you’ll just
make an ordinary, non-
static
method.

More by Author

Get the Free Newsletter!

Subscribe to Developer Insider for top news, trends & analysis

Must Read