Is it possible, in Java, to access fields of an object that is passed as a parameter in a method?
Example code:
void myMethod(ArrayList<Integer> list, MyClass object) {
Integer myInt = object.x; // x is an Integer-type field in the object
}
I tried the following:
MyClass curObj = (MyClass)object; to no avail.
Any suggestions?
When I use javac to compile, I get a cannot find symbol error.
If I understood the question correctly why simply not access object directly? e.g. if object had a member memberName and methodName() method, which are public, you could simply do
void myMethod(ArrayList<Integer> list, MyClass object) {
object.memberName = "member Name"
object.methodName();
}
You don't have to cast your 'object' as it is only object in name, you are getting passed a MyClass Object.
And by that I mean you are getting MyClass object not Object object
To access x like that make it sure it is public, BUT you should make an accessor method in MyClass:
public Integer getX()
{
return x;
}
and change your line to:
Integer myInt = object.getX();
No reasons it shouldn't work unless
object is null (but this would be a run-time, not compile-time error)
x is not public and myMethod is not a method of the MyClass class
Yes, it is possible if your attribute is declared as public. For example:
public class MyClass {
public int test;
}
class Test {
public printMyClass( MyClass c ) {
System.out.println( c.test );
}
}
Or you can have one method that returns test value and only MyClass can modify test attribute (better).
public class MyClass {
private int test;
public int getTest() {
return test;
}
}
class Test {
public printMyClass( MyClass c ) {
System.out.println( c.getTest() );
}
}
Only if MyClass.x is a public, then yes. Otherwise, you can reach it via that property's getter/setter methods. Java is actually passing a copy of the reference to the object, so while the called method is working only with the reference copy, the underlying object is the same whether you access it via a reference or a copy of the reference. The state of the object is also exposed for potential changes from within the called method.
Of course Your example code should work as is.
See that the field x has the word public in your MyClass.
//In MyClass
public Integer x;
If you don't declare it public you have to provide a getter method. This is the normal case (private + getter) JAVA Bean...
//In MyClass
private Integer x;
public Integer getX()
{
return x;
}
void myMethod(ArrayList<Integer> list, MyClass object) {
Integer myInt = object.getX(); // x is an Integer-type field in the object
}
Related
This question already has answers here:
When should I use "this" in a class?
(17 answers)
Closed 7 years ago.
I'm trying to get an understanding of what the the java keyword this actually does.
I've been reading Sun's documentation but I'm still fuzzy on what this actually does.
The this keyword is a reference to the current object.
class Foo
{
private int bar;
public Foo(int bar)
{
// the "this" keyword allows you to specify that
// you mean "this type" and reference the members
// of this type - in this instance it is allowing
// you to disambiguate between the private member
// "bar" and the parameter "bar" passed into the
// constructor
this.bar = bar;
}
}
Another way to think about it is that the this keyword is like a personal pronoun that you use to reference yourself. Other languages have different words for the same concept. VB uses Me and the Python convention (as Python does not use a keyword, simply an implicit parameter to each method) is to use self.
If you were to reference objects that are intrinsically yours you would say something like this:
My arm or my leg
Think of this as just a way for a type to say "my". So a psuedocode representation would look like this:
class Foo
{
private int bar;
public Foo(int bar)
{
my.bar = bar;
}
}
The keyword this can mean different things in different contexts, that's probably the source of your confusion.
It can be used as a object reference which refers to the instance the current method was called on: return this;
It can be used as a object reference which refers to the instance the current constructor is creating, e.g. to access hidden fields:
MyClass(String name)
{
this.name = name;
}
It can be used to invoke a different constructor of a a class from within a constructor:
MyClass()
{
this("default name");
}
It can be used to access enclosing instances from within a nested class:
public class MyClass
{
String name;
public class MyClass
{
String name;
public String getOuterName()
{
return MyClass.this.name;
}
}
}
"this" is a reference to the current object.
See details here
The keyword this is a reference to the current object. It's best explained with the following piece of code:
public class MyClass {
public void testingThis()
{
// You can access the stuff below by
// using this (although this is not mandatory)
System.out.println(this.myInt);
System.out.println(this.myStringMethod());
// Will print out:
// 100
// Hello World
}
int myInt = 100;
string myStringMethod()
{
return "Hello World";
}
}
It's not used a lot unless you have code standard at your place telling you to use the this keyword. There is one common use for it, and that's if you follow a code convention where you have parameter names that are the same as your class attributes:
public class ProperExample {
private int numberOfExamples;
public ProperExample(int numberOfExamples)
{
this.numberOfExamples = numberOfExamples;
}
}
One proper use of the this keyword is to chain constructors (making constructing object consistent throughout constructors):
public class Square {
public Square()
{
this(0, 0);
}
public Square(int x_and_y)
{
this(x_and_y, x_and_y);
}
public Square(int x, int y)
{
// finally do something with x and y
}
}
This keyword works the same way in e.g. C#.
An even better use of this
public class Blah implements Foo {
public Foo getFoo() {
return this;
}
}
It allows you to specifically "this" object in the current context. Another example:
public class Blah {
public void process(Foo foo) {
foo.setBar(this);
}
}
How else could you do these operations.
"this" keyword refers to current object due to which the method is under execution. It is also used to avoid ambiguity between local variable passed as a argument in a method and instance variable whenever instance variable and local variable has a same name.
Example ::
public class ThisDemo1
{
public static void main(String[] args)
{
A a1=new A(4,5);
}
}
class A
{
int num1;
int num2;
A(int num1)
{
this.num1=num1; //here "this" refers to instance variable num1.
//"this" avoids ambigutiy between local variable "num1" & instance variable "num1"
System.out.println("num1 :: "+(this.num1));
}
A(int num, int num2)
{
this(num); //here "this" calls 1 argument constructor within the same class.
this.num2=num2;
System.out.println("num2 :: "+(this.num2));
//Above line prints value of the instance variable num2.
}
}
The keyword 'this' refers to the current object's context. In many cases (as Andrew points out), you'll use an explicit this to make it clear that you're referring to the current object.
Also, from 'this and super':
*There are other uses for this. Sometimes, when you are writing an instance method, you need to pass the object that contains the method to a subroutine, as an actual parameter. In that case, you can use this as the actual parameter. For example, if you wanted to print out a string representation of the object, you could say "System.out.println(this);". Or you could assign the value of this to another variable in an assignment statement.
In fact, you can do anything with this that you could do with any other variable, except change its value.*
That site also refers to the related concept of 'super', which may prove to be helpful in understanding how these work with inheritance.
It's a reference of actual instance of a class inside a method of the same class.
coding
public class A{
int attr=10;
public int calc(){
return this.getA()+10;
}
/**
*get and set
**/
}//end class A
In calc() body, the software runs a method inside the object allocated currently.
How it's possible that the behaviour of the object can see itself? With the this keyword, exactly.
Really, the this keyword not requires a obligatory use (as super) because the JVM knows where call a method in the memory area, but in my opinion this make the code more readeable.
It can be also a way to access information on the current context.
For example:
public class OuterClass
{
public static void main(String[] args)
{
OuterClass oc = new OuterClass();
}
OuterClass()
{
InnerClass ic = new InnerClass(this);
}
class InnerClass
{
InnerClass(OuterClass oc)
{
System.out.println("Enclosing class: " + oc + " / " + oc.getClass());
System.out.println("This class: " + this + " / " + this.getClass());
System.out.println("Parent of this class: " + this.getClass().getEnclosingClass());
System.out.println("Other way to parent: " + OuterClass.this);
}
}
}
Think of it in terms of english, "this object" is the object you currently have.
WindowMaker foo = new WindowMaker(this);
For example, you are currently inside a class that extends from the JFrame and you want to pass a reference to the WindowMaker object for the JFrame so it can interact with the JFrame. You can pass a reference to the JFrame, by passing its reference to the object which is called "this".
Every object can access a reference to itself with keyword this (sometimes called the this
reference).
First lets take a look on code
public class Employee {
private int empId;
private String name;
public int getEmpId() {
return this.empId;
}
public String getName() {
return this.name;
}
public void setEmpId(int empId) {
this.empId = empId;
}
public void setName(String name) {
this.name = name;
}
}
In the above method getName() return instance variable name.
Now lets take another look of similar code is
public class Employee {
private int empId;
private String name;
public int getEmpId() {
return this.empId;
}
public String getName() {
String name="Yasir Shabbir";
return name;
}
public void setEmpId(int empId) {
this.empId = empId;
}
public void setName(String name) {
this.name = name;
}
public static void main(String []args){
Employee e=new Employee();
e.setName("Programmer of UOS");
System.out.println(e.getName());
}
}
Output
Yasir Shabbir
this operator always work with instance variable(Belong to Object)
not any class variable(Belong to Class)
this always refer to class non static attribute not any other parameter or local variable.
this always use in non static method
this operator cannot work on static variable(Class variable)
**NOTE:**It’s often a logic error when a method contains a parameter or local variable that has the
same name as a field of the class. In this case, use reference this if you wish to access the
field of the class—otherwise, the method parameter or local variable will be referenced.
What 'this' does is very simply. It holds the reference of current
object.
This keyword holds the reference of instance of current class
This keyword can not be used inside static function or static blocks
This keyword can be used to access shadowed variable of instance
This keyword can be used to pass current object as parameter in function calls
This keyword can be used to create constructor chain
Source: http://javaandme.com/core-java/this-word
Sometimes in a constructor, no statement is given. What does that indicate? For example if i create a class CIRCLE, then inside the class i write CIRCLE() {}, that is nothing is written inside. Can anyone explain it?
If your question is "why would anyone write such a constructor", then the answer is that the no-args default constructor only exists if no other constructor is specified.
Consider the following class.
class Foo {
int x;
}
As written, someone could write the following code to construct Foo.
Foo foo = new Foo();
However, now suppose I added a constructor which takes arguments.
class Foo {
int x;
public Foo(int x) {
this.x = x;
}
}
Now, suddenly, Foo foo = new Foo(); no longer works. To restore it, I must add the empty constructor again.
class Foo {
int x;
public Foo(int x) {
this.x = x;
}
public Foo() { }
}
Now, What if there are no other constructors that take arguments?
In that case, it is generally as the other answers suggest, to restrict access to constructing the class.
In the following definition of Foo, nobody is allowed to construct Foo. Perhaps Foo is meant only as a static class.
class Foo {
int x;
private Foo() { }
}
In the protected case, only subclasses can construct Foo.
class Foo {
int x;
protected Foo() { }
}
If there is no code in the constructor, chances are, it was declared to change the access to the constructor. By default, constructors are public. If you wanted to make it private, protected or package-private, you must explicitly declare it and manually change the modifier.
class Example {
public static void main(String[] args) {
new Demo(); //this is currently allowed
}
}
class Demo {
}
In order to prevent the creation of a Demo object within Example, we could declare Demo's constructor amd make it private:
class Demo {
private Demo() { }
}
Another reason could be that the class has a constructor that requires parameters. If so, you must explicitly declare the no-arg constructor to be able to use it.
If nothing is written, then when a new Object of that type is created, nothing 'extra' is done, whereas if in the constructor has code in, it does something.
For example, the following consructor for a class called 'Bank' assigns the argument 'name' to the field 'bankName', then instantiates a Terminal and 2 bank accounts:
private static final int INITIAL_BALANCE = 200;
public Bank( String name )
{
bankName = name;
atm = new Terminal();
account1 = new BankAccount( INITIAL_BALANCE );
account2 = new BankAccount( INITIAL_BALANCE );
}
It's a default constructor. For instance if you go:
Circle circle = new Circle();
You are then calling the default constructor. When you go ... Circle() that is a call to the default constructor, the one with no parameters.
The point of this is just to 'construct' an object or instantiate a class (instantiate just means create an object which is an instance of the class) with no additional information i.e. parameters.
This would generally be used to initialize fields to their default values, like so:
public Circle() {
this.x = 0;
this.y = 0;
}
I have an object, obj, of type MyObject, that I declare an instance of.
MyObject obj;
However, I don't initialize it. MyObject's Class looks something like:
public class MyObject {
public String i;
public String j;
public MyObject(String i) {
i = this.i;
}
}
So now, I want to set the value of j. So I say:
obj.j = "Hello";
Can I do this without having initialized obj? i.e. without saying:
obj = new MyObject("My i");
Will this object be null if I were to check the value of it, if I don't initialize it, or is setting a field within it enough to make it not null?
Thanks!
No, you cannot do that. You will have to create a new instance of MyObject if you want to access its fields.
Unless you make the fields static, ofcourse.
Do note that having your fields public violates encapsulation. You should make them private (or protected, if it's appropriate) and use getters and setters to provide access.
Sidenote:
public MyObject(String i) {
i = this.i;
}
This will not do what you want.
You have to assign the parameter i to the field variable i, not the other way around.
public MyObject(String i) {
this.i = i;
}
Please have a look at this code :
class Foo {
public int a;
public Foo() {
a = 3;
}
public void addFive() {
a += 5;
}
public int getA() {
System.out.println("we are here in base class!");
return a;
}
}
public class Polymorphism extends Foo{
public int a;
public Poylmorphism() {
a = 5;
}
public void addFive() {
System.out.println("we are here !" + a);
a += 5;
}
public int getA() {
System.out.println("we are here in sub class!");
return a;
}
public static void main(String [] main) {
Foo f = new Polymorphism();
f.addFive();
System.out.println(f.getA());
System.out.println(f.a);
}
}
Here we assign reference of object of class Polymorphism to variable of type Foo, classic polmorphism. Now we call method addFive which has been overridden in class Polymorphism. Then we print the variable value from a getter method which also has been overridden in class Polymorphism. So we get answer as 10. But when public variable a is SOP'ed we get answer 3!!
How did this happen? Even though reference variable type was Foo but it was referring to object of Polymorphism class. So why did accessing f.a not result into value of a in the class Polymorphism getting printed? Please help
You're hiding the a of Polymorphism - you should actually get a compiler warning for that. Therefore those are two distinct a fields. In contrast to methods fields cannot be virtual. Good practice is not to have public fields at all, but only methods for mutating private state (encapsulation).
If you want to make it virtual, you need to make it as a property with accessor methods (e.g. what you have: getA).
This is due to the fact that you can't override class varibles. When accessing a class variable, type of the reference, rather than the type of the object, is what decides what you will get.
If you remove the redeclaration of a in the subclass, then I assume that behaviour will be more as expected.
Right now I have two .java files.
The Main.java:
public class Main {
static int integer = 15;
NeedInteger need = new NeedInteger();
}
and the NeedInteger.java
public class NeedInteger {
System.out.println(integer);
}
This is of course very simplified, but is there any way I can accomplish this?
As many have answered, the correct method is to pass the value in to the constructor of the new class.
If for some reason you cannot do that, then you can use a public static accessor method in Main to access the value (this would be slightly better than just making the field public).
E.g.
public class Main
{
private static int integer = 15;
public static int getInteger()
{
return integer;
}
}
public class NeedInteger
{
public NeedInteger()
{
int integer = Main.getInteger();
}
}
Add a constructor to NeedInteger (and optionally a member if you need to also store it):
public class NeedInteger {
private int integer;
public NeedInteger(int integer) {
this.integer = integer;
System.out.println(integer);
}
}
Then pass your value when you create the instance:
public class Main {
static int integer = 15;
NeedInteger need = new NeedInteger(integer);
}
You would have to do some bad juju moves (like using a global variable) or pass it to the constructor.
NOTE: your
public class NeedInteger {
System.out.println(integer);
}
has no method in it. I would recommend all this to be rewritten as such:
public Class NeedInteger {
NeedInteger(int integer) {
System.out.println(integer);
}
}
If you really want the work to be done on construction.
EDIT: From your comment above.
Instead, have the class structured so:
public Class NeedStringArray {
NeedStringArray(String[][][] stringArr) {
//work with String array here
}
}
That has no real additional overhead, since the actual array will not be passed, but only a reference to it. You WILL likely want to set the array to be final or something, to avoid it being edited in the NeedStringArray constructors.
integer is private, so it cannot be accessed by NeedInteger. you'll have to make it public or use a setter or getter and you'll need to use Main.integer since it's static.
Generally, you set in the Constructor.
Pass in the variable to the class constructor.
An array reference would be just that--a reference.
Or you could pass in the class itself, or use a static (meh).
Per your comment I'd say you can either host your array in a singleton
or as others suggested have the second class accept the reference to the array in the constructor. You can then use Dependency Injection framework (e.g. Guice) to get wire them up