public class Curiosity {
public void toString()//error because of this specific method name
{
System.out.println("method is successfully implemented");
}
}
How can i use a method of the same name "toString()" if i want to ?
Do I have to give its return type as String if not what should i do to change its return type like suppose if i want to use a void return type for toString does java allow that ?
toString() method must return a String. That's the only way to override Object's toString().
public String toString()
{
return "method is successfully implemented";
}
If you wish to use the same name but not override Object's toString, you can overload the name toString by adding arguments, thus changing the signature of your method.
Example :
public void toString (String something)
{
System.out.println("method is successfully implemented " + something);
}
You are trying to overload toString() method in a wrong manner
Overloaded methods are differentiated by the number and the type of the arguments passed into the method. In the code sample, draw(String s) and draw(int i) are distinct and unique methods because they require different argument types.
You cannot declare more than one method with the same name and the same number and type of arguments, because the compiler cannot tell them apart.
The compiler does not consider return type when differentiating methods, so you cannot declare two methods with the same signature even if they have a different return type.
The only way to use toString() in your class is by keeping the return type as String
public String toString()
{
//your code here
}
That is how it is defined in Objectclass and if you wish to override it you will have to use the exact signature
or if you still wish to use the method name as toString what you can do is change the method's signature.
A method's signature includes method's name and the parameters.
Remember that return type is not a part of a method's signature
You can look at the source code of the java.lang.Object.
The toString method have a return value in String type. You can't have another method which's name is toString but return type is not String.
Actually, it's forbidden in Java in any inheritance relationship. When you call the method, the compiler only cares the name and the parameters. So how can it distinguishes the methods of the same name but with the different return type?
Related
I'm experimenting with this code:
interface Callee {
public void foo(Object o);
public void foo(String s);
public void foo(Integer i);
}
class CalleeImpl implements Callee
public void foo(Object o) {
logger.debug("foo(Object o)");
}
public void foo(String s) {
logger.debug("foo(\"" + s + "\")");
}
public void foo(Integer i) {
logger.debug("foo(" + i + ")");
}
}
Callee callee = new CalleeImpl();
Object i = new Integer(12);
Object s = "foobar";
Object o = new Object();
callee.foo(i);
callee.foo(s);
callee.foo(o);
This prints foo(Object o) three times. I expect the method selection to take in consideration the real (not the declared) parameter type. Am I missing something? Is there a way to modify this code so that it'll print foo(12), foo("foobar") and foo(Object o)?
I expect the method selection to take
in consideration the real (not the
declared) parameter type. Am I missing
something?
Yes. Your expectation is wrong. In Java, dynamic method dispatch happens only for the object the method is called on, not for the parameter types of overloaded methods.
Citing the Java Language Specification:
When a method is invoked (§15.12), the
number of actual arguments (and any
explicit type arguments) and the
compile-time types of the arguments
are used, at compile time, to
determine the signature of the method
that will be invoked (§15.12.2). If
the method that is to be invoked is an
instance method, the actual method to
be invoked will be determined at run
time, using dynamic method lookup
(§15.12.4).
As mentioned before overloading resolution is performed at compile time.
Java Puzzlers has a nice example for that:
Puzzle 46: The Case of the Confusing Constructor
This puzzle presents you with two Confusing constructors. The main method invokes a constructor,
but which one? The program's output depends on the answer. What does the program print, or is it
even legal?
public class Confusing {
private Confusing(Object o) {
System.out.println("Object");
}
private Confusing(double[] dArray) {
System.out.println("double array");
}
public static void main(String[] args) {
new Confusing(null);
}
}
Solution 46: Case of the Confusing Constructor
...
Java's overload resolution process operates in two phases. The first phase selects all the methods or constructors that are accessible and applicable. The second phase selects the most specific of the methods or constructors selected in the first phase. One method or constructor is less specific than another if it can accept any parameters passed to the other [JLS 15.12.2.5].
In our program, both constructors are accessible and applicable. The constructor
Confusing(Object) accepts any parameter passed to Confusing(double[]), so
Confusing(Object) is less specific. (Every double array is an Object, but not every Object is a double array.) The most specific constructor is therefore Confusing(double[]), which explains the program's output.
This behavior makes sense if you pass a value of type double[]; it is counterintuitive if you pass null. The key to understanding this puzzle is that the test for which method or constructor is most specific does not use the actual parameters: the parameters appearing in the invocation.
They are used only to determine which overloadings are applicable. Once the compiler determines which overloadings are applicable and accessible, it selects the most specific overloading, using only the formal parameters: the parameters appearing in the declaration.
To invoke the Confusing(Object) constructor with a null parameter, write new
Confusing((Object)null). This ensures that only Confusing(Object) is applicable. More
generally, to force the compiler to select a specific overloading, cast actual parameters to the declared types of the formal parameters.
Ability to dispatch a call to a method based on types of arguments is called multiple dispatch. In Java this is done with Visitor pattern.
However, since you're dealing with Integers and Strings, you cannot easily incorporate this pattern (you just cannot modify these classes). Thus, a giant switch on object run-time will be your weapon of choice.
In Java the method to call (as in which method signature to use) is determined at compile time, so it goes with the compile time type.
The typical pattern for working around this is to check the object type in the method with the Object signature and delegate to the method with a cast.
public void foo(Object o) {
if (o instanceof String) foo((String) o);
if (o instanceof Integer) foo((Integer) o);
logger.debug("foo(Object o)");
}
If you have many types and this is unmanageable, then method overloading is probably not the right approach, rather the public method should just take Object and implement some kind of strategy pattern to delegate the appropriate handling per object type.
I had a similar issue with calling the right constructor of a class called "Parameter" that could take several basic Java types such as String, Integer, Boolean, Long, etc. Given an array of Objects, I want to convert them into an array of my Parameter objects by calling the most-specific constructor for each Object in the input array. I also wanted to define the constructor Parameter(Object o) that would throw an IllegalArgumentException. I of course found this method being invoked for every Object in my array.
The solution I used was to look up the constructor via reflection...
public Parameter[] convertObjectsToParameters(Object[] objArray) {
Parameter[] paramArray = new Parameter[objArray.length];
int i = 0;
for (Object obj : objArray) {
try {
Constructor<Parameter> cons = Parameter.class.getConstructor(obj.getClass());
paramArray[i++] = cons.newInstance(obj);
} catch (Exception e) {
throw new IllegalArgumentException("This method can't handle objects of type: " + obj.getClass(), e);
}
}
return paramArray;
}
No ugly instanceof, switch statements, or visitor pattern required! :)
Java looks at the reference type when trying to determine which method to call. If you want to force your code you choose the 'right' method, you can declare your fields as instances of the specific type:
Integeri = new Integer(12);
String s = "foobar";
Object o = new Object();
You could also cast your params as the type of the param:
callee.foo(i);
callee.foo((String)s);
callee.foo(((Integer)o);
If there is an exact match between the number and types of arguments specified in the method call and the method signature of an overloaded method then that is the method that will be invoked. You are using Object references, so java decides at compile time that for Object param, there is a method which accepts directly Object. So it called that method 3 times.
Groovyc: [Static type checking] - Cannot find matching method io.swagger.client.util.EmUtil#addLobList(java.lang.Object, java.lang.Object). Please check if the declared type is correct and if the method exists.
I am calling this function in a static block in groovy, as shown below:
static {
Arrays.asList(LOBEnum.values()).forEach {lob -> EmUtil.getInstance().addLobList(lob.name(), lob.getLob())}
EmUtil.getInstance().setPrefix("CCB_Reference_Data_")
EmUtil.getInstance().init()
}
This calls a java function, as shown below:
public void addLobList(String lob, String licenseLob) {
lobList.add(lob);
lobLicenseList.add(licenseLob);
}
It looks like lob.name() and lob.getLob() each have a return type of Object. If you want to use the return values of those methods as parameters to your addLobList method, the static type checker needs to make sure those are String instance. You can cast those with (String). Alternatively, you could change the return value of those methods to String.
This question already has answers here:
Can someone explain a void return type in Java?
(5 answers)
Closed 6 years ago.
I'm confused about "void",
as it pertains to methods.
I don't know what the distinction between two methods is when one has "void" and another doesn't.
For example, if I do:
Public meth (int amount)
{
amount = initial * interest;
return amount;
}
( not sure if it was right, or even valid, to take the name "amount" and name it the same thing as my formal parameter, but what makes sense here is that you're performing a calculation and returning the result)
Then, if I did something like:
Public void Testing (int array[])
{
//code that would modify the internals of an array
}
Would the second one have no "return" because it's more of a general method, that can be applied to any integer array, while the first one is about doing work on specific variables?
Would also appreciate one or two more examples of when I would or wouldn't be using "void" and "return".
One other thing that seems to confuse me is calling methods.
I know sometimes I'll do something like, for example, using the Testing method above,
Testing(ArrayName);
Other times, it will be like:
NameOfWhateverImApplyingMethodTo.MethodName();
And then there are times when things will be done properly by:
Thing1.MethodName(Thing2);
Which circumstances would I switch the syntax for method calls like this?
Java is case sensitive, so the modifier Public is invalid, use public
You can't define a method as public methodName(int a), only a constructor has this signature, a method must be public void methodName(<signature>) for methods that don't return anything or public <return type> methodName(<signature>) for methods that do.
Void basically means that the method will not return anything.
If you did
String name= "tim";
public void getName(){
return name;
}
This would result in an error, because the getName method is returning a string object called name, but the method declaration is saying I am returning nothing - because it is void.
Instead the method should be :
String name = "tim";
public String getName(){
return name;
}
Now when the method getName() is called it will return a string object "name" with "tim" inside of it :)
You might have void for a set method. So for example
String name = "tim";
public void setName(String newName){
this.name = newName;
}
When this method is called you would use setName("Andy"); and it would set the value of the name variable to be "Andy". Nothing is returned in this method, because it is setting something, but there is no need to send anything back, so we use void on the method declaration.
Hope this helps.
The method that has void as return type does not return anything. For example you want to set a field firstName in your class. You will write a setting method like
public void setFirstName(String n) {
this.firstName = n;
}
As you can see you are just setting a class variable and does not require to return anything.
If you dont use void then you have to provide a return type for method. Like if you wish to write a getter for above variable as:
public String getFirstName() {
return this.firstName;
}
Once you provide a return type, you will have to return a value of that type otherwise your code will not compile.
Calling a method can be done based on where you are calling it from and what modifier is used:
If you are calling the method from the same class then you can simply write firstName = getFirstName()
If you are calling the method from another class then you require object of method's class as qualifier like personObject.getFirstName()
If you are calling a static method then you require class name as qualifier like Person.getFirstName();
Return type is what you get out of it. When you call it, what are you hoping to get back? For instance, if the method gets the average of two numbers, then you're expecting a number back, so the return type will be a number type, like "int" (integer).
You can see what it should be using that logic or by looking in the method for the word return - what comes after return is what is returned, and its type should be declared in the method (e.g. if it says "return 4;" it's returning an int, and should be e.g. public int getFour()
You also asked about e.g. testing() vs testing(word)
I remember having the same difficulty. The distinction between the two also relates to the method declaration line. I'll illustrate.
public String testing(){
return "a word";
}
Calling this method by doing "System.out.println(testing());" should print "a word". Calling this method by doing "System.out.println(testing("a word"));" will give you an issue - this is because when you call testing, it looks at the appropriate method: one in the right class, with the right return type and with the right arguments/parameters. If you're calling testing("a word"), that means you're using a String as an argument (because "a word" is a string), and so it tries to use the testing(String aString) method - which doesn't exist.
So you use empty brackets when the method takes no input, and you put stuff in brackets when the method expects stuff. This should be less confusing than it sounds, because it's usually logical - if you want to call a method that returns an average, you need to ask yourself "Average of what?" You'd probably need to supply it with the values you want the average of.
Moving on: (a) testing() versus(b) AClass.testing() versus(c) aclass.testing() -
In (a), there's no class specified. Therefore, if you call it from that class, Java can guess which class: this one, and it'll work. From any other class, it won't know what you're talking about, and might even insult you.
In (b), you're specifying a class in general - therefore it'll know what class to find it in - and it'll work if it's a "static method". *[see bottom]
In (c), you're specifying an instance of AClass you want to run "testing()" on*.
For instance, imagine you've created a class called Business. You make a hundred Business objects by specifying for each a name, number, address.
e.g.
Business b = new Business(name, number, address);
Then in the Business class you have a method "getName()". This method takes no argument - you could see that the brackets are empty - so if, from another class, you call "Business.getName()", how could it know which name you want? You've just made a hundred businesses!
It simply can't. Therefore, for such a method, you'd call "b.getName()" (b being the Business we created above) and it would get the name for this instance of a Business - namely, b.
I'm happy to help, so if you're confused about any particular parts of what I just wrote please let me know and I'll try to elaborate!
edit: A bit on static methods:
Static methods don't belong to an instance of the class. getName(), for example, would get the name of this Business - ie, this instance of the Business class. But let's say that in the Business class you made a method that took the first letter of each word in a String and transformed it to uppercase - like if you wanted to make the business names look more professional when you printed them out.
public static String stringToUpperCase(String aString){
aString = aString.substring(0, 1).toUpperCase() + aString.substring(1);
return aString;
}
And to use that, you change the getName() method from:
public String getName(){
return name;
}
to
public String getName(){
return stringToUpperCase(name);
}
The new method is used here to make the name have an uppercase first letter - but that is the extent of its involvement with the Business class. You notice it doesn't ask for information about the name, address, or number for a particular business. It just takes a string you give it, does something to it, and gives it back. It doesn't matter whether you have no Businesses or a hundred.
To call this method, you'd use:
System.out.println(Business.stringToUpperCase("hello"));
This would print Hello.
If it were not a static method, you'd have to make a new Business first:
Business b = new Business("aName", "aNumber", "anAddress");
System.out.println(b.stringToUpperCase("hello"));
And if the method did need access to more Business-instance information (like a business's name number or address) it wouldn't be able to be an instance variable.
The first example, a method without a return type at all, is a constructor; used when an instance is created with new. However, you can't return a value from a constructor. Something like,
this.amount = initial * interest; // return amount;
Sets the field amount to initial * interest.
This does not work
public String getjLabel4text(void){...
But this does
public String getjLabel4text(){...
Why so? I am not taking any arguments so shouldn't I write void there? Why is it causing an error?
More generally, method declarations have six components, in order:
Modifiers—such as public, private, etc.
The return type—the data type of the value returned by the method, or void if the method does not return a value.
The method name—the rules for field names apply to method names as well, but the convention is a little different.
The parameter list in parenthesis—a comma-delimited list of input parameters, preceded by their data types, enclosed by parentheses, (). If there are no parameters, you must use empty parentheses.
An exception list—to be discussed later.
The method body, enclosed between braces—the method's code, including the declaration of local variables, goes here.
Oracle Documentation
Java is not like C/C++ You do not need to put void parameter on a function with no parameters, and it throws an error because it is not a valid parameter definition as in java you can not have parameters of type void.
In C and C++ we can take void as an method argument because void is a datatype in both c and c++ . But Java is different, Here void is not a datatype rather its a keyword that is used just to signify method will not return any argument. so taking void as argument is technically invalid and is of no use, this () itself is sufficient.
Ex.
public String Name() { return "Hello";}
and
public String Name(void){ return "Hello";}
In C : Name() will take unspecified no of argument for unspecified type and Name(void) will take no argument.
In C++ : Both Name() and Name(void) will take no argument.
In Java : Name() will work while Name(void) will return compilation error.
There is no need to explicitly mark method as no argument in Java.
You just omit any argument definition and there it is, a method with no arguments.
There is, though a need for return type for any method. It's where void gets in.
void is just for specifying a return value. Declaration of a return type is required for the syntax to work correctly. Since the argument list length is variable, having none is acceptable. void is a keyword, and is not allowed in the argument list.
A method can return a value or not.
public void doSomthing(String text) {
print text;
}
This don't return any value but executes and returns.
public String doSomthing(String text) {
return text;
}
This method returns the text.
You can't input a void.
If I am printing an object of the class then it is printing the toString() method implementation even I am not writing the toString() method so what are the implementation,how it is calling toString() internally?
You're not explicitly calling toString(), but implicitly you are:
See:
System.out.println(foo); // foo is a non primitive variable
System is a class, with a static field out, of type PrintStream. So you're calling the println(Object) method of a PrintStream.
It is implemented like this:
public void println(Object x) {
String s = String.valueOf(x);
synchronized (this) {
print(s);
newLine();
}
}
As we see, it's calling the String.valueOf(Object) method.
This is implemented as follows:
public static String valueOf(Object obj) {
return (obj == null) ? "null" : obj.toString();
}
And here you see, that toString() is called.
Every object in Java IS-A(n) Object as well. Hence, if a toString() implementation has not been provided by a class the default Object.toString() gets invoked automatically.
Object.toString()'s default implementation simply prints the object's class name followed by the object's hash code which isn't very helpful. So, one should usually override toString() to provide a more meaningful String representation of an object's runtime state.
even I am not writing the toString() method so what are the implementation,how it is calling toString() internally?
toString() is one of the few methods (like equals(), hashCode() etc.) that gets called implicitly under certain programmatic situations like (just naming a few)
printing an object using println()
printing a Collection of objects (toString() is invoked on all the elements)
concatenation with a String (like strObj = "My obj as string is " + myObj;)
Everything inherits from Object, so the toString on Object will be called if you have not defined one.
toString() method is present in Object class, so when u put obj in System.out.println(obj);, impliciyly it will call toString() present in Object class since every user created class will implicitly inherits Object class so as ur newly created class, that means that toString() is available in ur class so it will print something like for example: "PkgNamePackage.Classname#12cf4"
However if u explicitely override toString method and give ur own implementation then it will written the string what ever u give in Overriden tostring method();
ex:
public class DogArray {
#Override
public String toString() {
return "Im the newly created Object";
}
public static void main(String args[]) {
DogArray d1 = new DogArray();
System.out.println(d1);
}
}
output: Im the newly created Object
In java object class is super class to the each and every class.whenever your passing parameter to the system.out.println internally object class to string method will be excuted.it returns class name#reference value given but as per our application requirement object class to string method will override in collection and string class.it returns their content.