This question already has answers here:
How to avoid constructor code redundancy in Java?
(4 answers)
Closed 9 years ago.
Hi I am just learning about constructor chaining in Java and had some questions...
First of all could someone please explain when I would ever need to use this? Off the top of my head I seriously cannot think of a situation.
In this example, within the constructor with no arguments I call another constructor. How do I access this new "James Bond" object for future use?
import java.util.*;
class Employee
{
private String name;
private double salary;
public Employee()
{
this("James Bond", 34000);
}
public Employee(String n, double s)
{
name = n;
salary = s;
}
public String getName()
{
return name;
}
public double getSalary()
{
return salary;
}
public static void main(String[] args)
{
Employee a = new Employee();
}
}
Actually I believe the most common use of chained Constructors is when the Constructor does more than just setting the member variables.
static int numOfExamples = 0;
public Example(String name, int num)
{
this.name = name;
this.num = num;
numOfExamples++;
System.out.println("Constructor called.");
Log.info("Constructor called");
}
public Example()
{
this("James Bond",3);
}
That way we don't have to write the code for logging and incrementing the static variable twice, instead just chaining the constructors.
Chaining constructors like this is useful to avoid repeating code, and helps with maintainability:
public MyClass(int x, double y, String z) {
// set fields
}
public MyClass() { // i.e. a constructor that uses default values
this(42, 4.2, "hello world"); // x is 42, y is 4.2, and z is "hello world"
}
If we didn't use the chain, and wanted to change how the x argument (for example) is processed when constructing an instance of MyClass, we would have to change code in both constructors. With the chain, we only need to change one of them.
1) As others have said, it's for code maintenance, basically the idea is that you only write one piece of code once, which means you only need to edit it once, there is no risk of overlooking something when editing your methods and the two accidentally becoming different.
Personally I tend to use this differently than in your example. Like so:
Employee() {
setupStuff();
}
Employee(String name) {
this();
this.setName(name);
}
This is a nice way of doing things, because potentially your setter can be way more complicated than just setting a member in the class. So basically what this does is puts calling the empty constructor and then a setter into a single method, making it much easier for anyone using the class.
2) The constructor being called doesnt't create a different object at all, it creates this object. Note that there is no new keyword used. Basically you're just calling a different method inside your constructor, except that method happens to also be a constructor.
Every time you want to allow constructing an object wit default values, and also want to allow creating the same object with non-default values. Let's imagine a DateTime class. You could want to initialize it with the current time by default, or with a specific time. Imagine a Car class. You could imagine constructing it with Black as the default color, or with a specific color. This kind of situation is very common. See java.util.ArrayList or java.util.Locale, for concrete examples.
It's stored in the name field. So you access it, from the object itself, with this.name (this being optional, just as in the getName() method).
How do I access this new "James Bond" object for future use?
Because you saved the values of name and salary as fields of your employee class, then inside the employee class you can use those fields, and outside your employee class you can use the getter/setter methos of your employee class
Related
I have a Java class that has a few members. I want to write a custom cast for it. I was wondering how is it possible to do so?
Let's assume the class is as follows:
class Person {
private int age;
private float weight;
// getters and setters and etc
}
I would like the int cast to return the member age of an object, and the float cast to return the weight of an object.
For instance:
public class Main {
public static void main(String[] args) {
// create an object
Person P = new Person();
P.setAge(21);
P.setWeight(88.0);
// case 1: casting object to an existing data type
int personAge = (int) P; // would return the age
float personWeight = (float) P; // would return the weight
// case 2: casting an existing data type to an object
Person P2 = (Person) personAge; // this would create an instance of the object whose age is assigned and weight is not assigned
}
}
I was wondering if it is possible to do the opposite. In particular, casting int to Person would return an instance of Person that its age is assigned and similarly for float.
I know this question may not have an answer. But because I did not find any useful results in my search, I decided to ask it.
P.S. I understand that for a String, the toString method would take care of case 1.
You can't overload the cast operator. Java doesn't support it and probably never will.
To convert a single value to an instance of the desired class, we use static factory methods.
public static Person fromAge(int age) {
return new Person(age);
}
They often return a partially constructed object. In the snippet above, a newly constructed person has only age set: other fields will have their default values.
To do the opposite, we use getters.
public int getAge() {
return age;
}
However, since toString is already there, it makes sense to add other data types as well.
toInt makes no sense when it's applied to me (as an instance of the Person class). It could be my height, my weight, my age, a number of times I went to a bathroom today, etc. I can't represent myself by one int number, neither can a large majority of classes.
On the other hand, toString can do this job pretty well: I can give you (read return) a summary of my hobbies, my biometric information, even my picture. Or I can leave it to the default implementation, which still would satisfactorily represent an object.
You wouldn't use a cast for this just write methods in your Person class to get those values.
public int getAge()
{
return age;
}
etc.
So I've only done this once and I'm unsure whether the approach I did is the appropriate way.
But my approach was a method called typeConverter, which i would give an object as parameter, then you could take that parameter and look what object type it is and then create a new Person, with your value.
Although this approach could cause problems, when your class would have two integer fields. But I think you could find a solution for this, by giving it another parameter that defines, which field you'd want to convert it to.
I'm really sorry for my poor english, but I hope you get the principle.
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.
Is there a name for this techinique (method calls returning objects upon which another method call is made on the same line)?
String pAID = commonAssets.getApplicationServerSettings().getSetting("plivoAuthID");
instead of
ApplicationServerSettings applicationServerSettings = commonAssets.getApplicationServerSettings();
String pAID = applicationServerSettings.getSetting("plivoAuthID");
Also, when I do the first, Eclipse doesn't prompt me to import the class ApplicationServerSettings, but it does if I use the second code style.
Also, are these two styles merely preferences?
The technique is called method chaining.
String pAID = commonAssets.getApplicationServerSettings().getSetting("plivoAuthID");
Definition from the wiki:
Method chaining, also known as named parameter idiom, is a common
syntax for invoking multiple method calls in object-oriented
programming languages. Each method returns an object, allowing the
calls to be chained together in a single statement without requiring
variables to store the intermediate results.[1] Local variable
declarations are syntactic sugar because of the difficulty humans have
with deeply nested method calls.[2][3] A method chain is also known as
a train wreck due to the increase in the number of methods that come
one after another in the same line that occurs as more methods are
chained together[4] even though line breaks are often added between
methods.
Your second question:
Also, when I do the first, Eclipse doesn't prompt me to import the class ApplicationServerSettings, but it does if I use the second code style.
From the definition again "Each method returns an object, allowing the calls to be chained together in a single statement without requiring variables to store the intermediate results." That is why it does not prompt you to import the class ApplicationServerSettings.
Another example (besides the want you introduce) that looks simpler:
Take a look at the wiki example:
class Person {
private String name;
private int age;
// In addition to having the side-effect of setting the attributes in question,
// the setters return "this" (the current Person object) to allow for further chained method calls.
public Person setName(String name) {
this.name = name;
return this;
}
public Person setAge(int age) {
this.age = age;
return this;
}
public void introduce() {
System.out.println("Hello, my name is " + name + " and I am " + age + " years old.");
}
// Usage:
public static void main(String[] args) {
Person person = new Person();
// Output: Hello, my name is Peter and I am 21 years old.
person.setName("Peter").setAge(21).introduce();
}
}
It's often called fluent syntax.
IMHO it's a matter of style, there's no right or wrong.
Fluent syntax is more concise, which is sometimes a good thing.
The other variant is more handy for source-level debugging. You can step through the statements and inspect the intermediate results.
Ok I am trying to wrap my head around this:
Write an application that creates a class for the student object with the following attributes:
Student Number, Name, Address, Phone No., and Course.
Write a test program that sets and gets each attribute in the class.
The test program should also display all of the attributes in the class.
Using the student class and its attributes from the previous question, write an application (extend the previous program) that includes both a user-defined and default constructor.
Write a test program to demonstrate the use of both constructors.
This is a work-sheet from college, for some revision on Objects in Java.
The part that troubles me is the one where it asks to have both user-defined and default constructor?
I was under impression it was impossible to do it? Because if you don't provide a constructor yourself, JVM will provide one (default constructor?). But if you define any constructor, then default one becomes unavailable?
Is this just poorly worded task, or could it mean something else?
I'm pretty sure whoever created the work-sheet meant "No-arg constructor" instead of "Default constructor".
You can't technically create the Default constructor (since this is done for you by the compiler), but you can explicitly create the no-arg constructor (which functionally is the same).
So, yes, poorly worded task.
In java, when you don't explicitly specify a constructor, the compiler will add a "default" constructor: a constructor thar doesn't take parameters. If you specify a constructor, then the compiler doesn't add that constructor.
For instance, this code will compile fine:
class Student {
String name;
int age;
// ...
}
// ...
Student myself = new Student();
But this code wont compile:
class Student {
String name;
int age;
// ...
public Student(String name) {
this.name = name;
}
}
// ...
Student myself = new Student(); // compilation error: use new Student("Jhon Smith");
Because the default constructor is not available any more.
Java provide default constructor when you dont implement yourselve one. But when you create customized constructor you have to implement also default if you would like to use constructor with no arguments. Let's asume we have class A{} for java it will looks like that:
public class A{
public A(){//auto generated constructor
}
}
but if you provide an customized constructor auto generated constructor dissapear.
Default constructor is not created when programmer provides any constructor. But here I'm expecting that the author of this task understands "default" constructor as the one without any parameters.
Concluding you would have two constructors:
public class MyClass {
public MyClass () {
}
public MyClass (long studentNumber, String name, String address....) {
}
Above is correct, however, in OO terms, "default constructor" is a constructor that takes in no arguments. The other type of constructor is one where arguments are taken into the constructor to make things custom.
ie:
Student(){ //default constructor
number = 0;
name = "bob";
//etc etc
}
student(int nm, int nm, etc etc){ //parametrized constructor
number = nm;
name = nm;
//etc etc
}
According to Charatan & Kans, Java in Two semesters 3rd Edition, page 196:
Constructor just like any methods can be overloaded, meaning within a single class we can have 2 or more constructors, where one take 1 argument like:("studentName") and within the same class, another constructor might take 2 arguments ("studentName", " studentId"), yet another constructor still within the same class may have 3 arguments ("studentName", "studentId", " studentPhoneNumber").
This is Constructor Overloading. Here's an example:
public class student {
// attributes
private String studentName;
private int studentID;
private int studentPhoneNumber;
// constructor with one argument
public student (String studentNameIn) {
studentName = studentNameIn;
}
// constructor with 2 arguments
public student (String studentNameIn, int studentIdIn) {
studentName = studentNameIn;
studentID = studentIdIn;
}
// constructor with 3 arguments
public student (String studentNameIn, int studentIdIn, int studentPhoneNumberIn) {
studentName = studentNameIn;
studentID = studentIdIn;
studentPhoneNumber = studentPhoneNumberIn;
}
// default constructor REINSERTED no argument
public student () {}
// methods
}
The above is a clear example of constructor overloading. Now during the OBJECT creation, meaning when the student object is being created, it will be up to the programmer to utilise the constructor he/she chooses, with 0 arguments ( default that was REINSERTED) or with 1 argument again he/she may choose the constructor that contains 2 arguments etc. Its a matter of choice and requirement of user.
No superclass or multiple class is needed as constructor can be overloaded as demonstrated above.
I am totally confused with ArrayList behavior. Wrote really long post, then realized no one is going to analyse huge code, so just core of the problem. Numbers are for convenience, but in my app these 0 and 24 are dynamic values.
ArrayList<VoipBlock> sortedBlocks = new ArrayList<VoipBlock>();
VoipBlock vb3 =new VoipBlock();
vb3=sortedBlocks.get(0);
vb3.setPacketNumber(24);
Essentially my final aim is to: modify and add back to arrayList as new value. However when I do that the guy at position 0 in ArrayList -> unsortedBlocks.get(0); replicates all the changes done to vb3 which of course is not what I want. I want vb3 acquire same values as VoipBlock inside of ArrayList, but I want it to be detached.
This is yet another case of passing by reference. I hate technical explanations - Java passes everything by value, BUT in some cases it passes references by values - this is same as saying not-oily oil. Please help.
It reminds me my start of learning JavaScript - I hated the language - until I watched proper materials at lynda.com - JavaScript Good Practices? - Diagrams killed me. It is the lazy description that turns us-youth away from brilliant technology, not the technology itself.
Please don't let it bother my stress and don't be in any way offended by me, it is just general complaining, maybe someone will look at it and make life better :-)
Thanks for Your time,
Desperately awaiting for help :-)
To achieve your objective you can use clone method. you have to override this method in VoipBlock class
Lets say VoipBlock is as follows
public class VoipBlock {
private int packetNumber;
private String type;
public int getPacketNumber() {
return packetNumber;
}
public String getType() {
return type;
}
public void setPacketNumber(int value) {
packetNumber = value;
}
public void setType(String value) {
type = value
}
public VoipBlock clone() {
VoipBlock clone = VoipBlock();
clone.setType(this.getType());
clone.setPacketNumber(this.getPacketNumber());
return clone;
}
}
So, using the same code you can do like as follows
ArrayList<VoipBlock> sortedBlocks = new ArrayList<VoipBlock>();
VoipBlock vb3 =new VoipBlock();
sortedBlocks.add(vb3);
vb3=sortedBlocks.get(0).clone();
vb3.setPacketNumber(24);
Note that upon calling clone method in above code segment, vb3 get assigned with a new VoipBlock instance. And already inserted VoipBlock to the array remains unchanged.
if you are looking to have kind of sample instances of VoipBlock instances which you later wanted to use in creating similar instances like them. check on immutability/mutability aspect of the code. check "Effective Java" by Joshua Blouch
The following will always copy the reference of b to a:
AnyClass a = ...;
AnyClass b = ...;
a = b;
What you want is probably to clone the object:
a = b.clone();
If I understand correctly, you're a bit unsure about how references and values work. I think the rule of thumb is that primitive types like int, char, boolean and maybe String are copied but Objects just have their reference passed.
The line vb3=sortedBlocks.get(0); completely replaces whatever vb3 used to be with the first thing in the ArrayList. And yes, it won't be a copy, it will be a reference to the same object in memory. So whatever you do will affect both of them. You need to either manually copy over all the information you need or to use a clone() or copy() function.
So for example, in your code, the line VoipBlock vb3 =new VoipBlock(); is a bit redundant because you're overwriting the new instance straight away.
What you really need here is to either use a copy constructor or declare VoipBlock to be Clonable so you can use the clone() method.
What you are interpreting as passing by reference is not actually passing by reference. Java objects are really pointers. Because of this you are passing the value of the pointer. So when you do:
vb3=sortedBlocks.get(0);
you are really assigning vb3 to point to the same locations in memory as sortedBlocks.get(0). Therefore when you manipulate vb3 properties through their setters, the result is seen in both.
If you want two separate pointers you need to use the new keyword or use the clone() method which does this under the hood.
An example to prove this is:
public class Person {
private String name;
public Person(String name) {
this.name = name;
}
public String getName() {
return this.name;
}
public void setName(String name) {
this.name = name;
}
}
public class Main {
public void doSomething(Person p) {
p = new Person("Bob");
System.out.println(p.getName());
}
public static void main(String[] args) {
Person p = new Person("Billy");
System.out.println(p.getName());
doSomething(p);
System.out.println(p.getName());
}
}
Since Java is pass by value the output will be: Billy, Bob, Billy. If Java were pass by reference it would be Billy, Bob, Bob. If I did not do the new Person in the doSomething() method and instead used the setName() method I would end up with Billy, Bob, Bob also but this is due to the fact I'm now modifying off the same pointer not that I passed by reference as the example above proves that's not the case.