How is the 'this' variable in Java actually set to the current object?

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囚心锁ツ
囚心锁ツ 2021-02-03 21:19

Consider:

class TestParent{
  public int i = 100;
  public void printName(){
    System.err.println(this); //{TestChild@428} according to the Debugger.
    Syste         


        
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  • 2021-02-03 21:49

    Well when a new object is created that object has an address in memory so you can think of it as if the object had a private member this that is set to the address when the object is created. You can also think of it like this: obj.method(param) is just syntactic sugar for method(obj, param); and this is actually a parameter of method.

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  • 2021-02-03 22:04

    What's happening here is that there are two completely different fields both called i; to use their full names, one is TestParent::i and one is TestChild::i.

    Because the method printName is defined in TestParent, when it refers to i, it can only see TestParent::i, which is set to 100.

    Whereas when you set i to 200 in TestChild, both fields called i are visible, but because they have the same name, TestChild::i hides TestParent::i, and you end up setting TestChild::i and leaving TestParent::i untouched.

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  • 2021-02-03 22:05

    In essence, there is no difference between

    this.foo()
    

    and

    anyObject.foo()
    

    as both are "implemented" the same way. Keep in mind that "in the end" "object orientation is only an abstraction, and in "reality" what happens is something like:

    foo(callingObject)
    

    In other words: whenever you use some object reference to call a method ... in the end there isn't a call on some object. Because deep down in assembler and machine code, something like "a call on something" doesn't exist.

    What really happens is a call to a function; and the first (implicit/invisible on the source code level) parameter is that object.

    BTW: you can actually write that down in Java like:

    class Bar {
       void foo(Bar this) { ... }
    

    and later use

    new Bar().foo();
    

    And for this.fieldA, in the end: you have a reference to some location in memory; and a table that tells you on which "offset" you will find fieldA.

    Edit - just for the record. If you are interested in more details about foo(Bar this) - you can turn to this question; giving the details in the Java spec behind it!

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  • 2021-02-03 22:09

    To directly address what you see in the output: The call to print 'this.i' is passing as argument to 'print()' the value of the field 'i' in the current scope, which is the scope of the parent class. By contrast, the call to print 'this' is getting translated under the hood to a call to print 'this.getClass().getName()' [roughly speaking], and the 'getClass()' call gets the actual class object, which is for the child class.

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  • 2021-02-03 22:09

    Adding some more info on top of @Tom Anderson answer, which explains hiding concept nicely.

    I have added one more constructor in Child ( TestChild) which prints values of i in both parent and child.

    If you want get value of i from child (TestChild), override the method in TestChild.

    class TestParent{
      public int i = 100;
      public void printName(){
        System.err.println("TestParent:printName()");
        System.err.println(this); //{TestChild@SOME_NUM} according to the Debugger.
        System.err.println(this.i); //this.i is 100.
      }
    }
    
    class TestChild extends TestParent{
      public int i = 200;
      public TestChild(){
        System.out.println("TestChild.i and TestParent.i:"+this.i+":"+super.i);
      }
      public void printName(){
          //super.printName();
          System.err.println("TestChild:printName()");
          System.err.println(this); //{TestChild@SOME_NUM} according to the Debugger.
          System.err.println(this.i); //this.i is 200.
      }
    }
    
    public class ThisTest {
      public static void main(String[] args) {
        TestParent parent = new TestChild();
        parent.printName();
      }
    }
    

    Case 1: If I comment super.printName() call from child, Child version of TestChild.printName() prints the value of i in TestChild

    Output:

    TestChild.i and TestParent.i:200:100
    TestChild:printName()
    TestChild@43cda81e
    200
    

    Case 2: TestChild.printName() calls super.printName() as the first line in printName() method. In this case, i value from both parent and child are displayed in respective methods.

    Output:

    TestChild.i and TestParent.i:200:100
    TestParent:printName()
    TestChild@43cda81e
    100
    TestChild:printName()
    TestChild@43cda81e
    200
    
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