visual basic barcode program Assigning One Reference Variable to Another in Java

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Assigning One Reference Variable to Another
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With primitive variables, an assignment of one variable to another means the contents (bit pattern) of one variable are copied into another. Object reference variables work exactly the same way. The contents of a reference variable are a bit pattern, so if you assign reference variable a to reference variable b, the bit pattern in a is copied and the new copy is placed into b. If we assign an existing instance of an object to a new reference variable, then two reference variables will hold the same bit pattern a bit pattern referring to a specific object on the heap. Look at the following code:
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import java.awt.Dimension; class ReferenceTest { public static void main (String [] args) { Dimension a = new Dimension(5,10); System.out.println("a.height = " + a.height); Dimension b = a; b.height = 30; System.out.println("a.height = " + a.height + "after change to b"); } }
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In the preceding example, a Dimension object a is declared and initialized with a width of 5 and a height of 10. Next, Dimension b is declared, and assigned the value of a. At this point, both variables (a and b) hold identical values, because the contents of a were copied into b. There is still only one Dimension object the one that both a and b refer to. Finally, the height property is changed using the b
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3: Operators and Assignments
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reference. Now think for a minute: Is this going to change the height property of a as well Let s see what the output will be:
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%java ReferenceTest a.height = 10 a.height = 30 after change to b
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From this output, we can conclude that both variables refer to the same instance of the Dimension object. When we made a change to b, the height property was also changed for a. One exception to the way object references are assigned is String. In Java, String objects are given special treatment. For one thing, String objects are immutable; you can t change the value of a String object. But it sure looks as though you can. Examine the following code:
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class Strings { public static void main(String [] args) { String x = "Java"; // Assign a value to x String y = x; // Now y and x refer to the same String object System.out.println("y string = " + y); x = x + " Bean"; // Now modify the object using the x reference System.out.println("y string = " + y); } }
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You might think String y will contain the characters Java Bean after the variable x is changed, because strings are objects. Let s see what the output is:
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%java String y string = Java y string = Java
As you can see, even though y is a reference variable to the same object that x refers to, when we change x it doesn t change y! For any other object type, where two references refer to the same object, if either reference is used to modify the object, both references will see the change because there is still only a single object. But with a string, the VM creates a brand new String object every time we use the + operator to concatenate two strings, or any time we make any changes at all to a string. You need to understand what happens when you use a String reference variable to modify a string:
A new string is created, leaving the original String object untouched. The reference used to modify the String (or rather, make a new String by
modifying a copy of the original) is then assigned the brand new String object.
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So when you say,
1. String s = "Fred"; 2. String t = s; // Now t and s refer to the same String object 3. t.toUpperCase(); // Invoke a String method that changes the String
you actually haven t changed the original String object created on line 1. When line 2 completes, both t and s reference the same String object. But when line 3 runs, rather than modifying the object referred to by t (which is the one and only String object up to this point), a brand new String object is created. And then abandoned. Because the new String isn t assigned to a String variable, the newly created String (which holds the string FRED ) is toast. So while two String objects were created in the preceding code, only one is actually referenced, and both t and s refer to it. The behavior of strings is extremely important in the exam, so we ll cover it in much more detail in 6.
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