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String objects and their reference variables
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String x = "Java"; x.concat(" Rules!"); System.out.println("x = " + x);
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6: Java.lang The Math Class, Strings, and Wrappers
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The output will be x = Java. The first line is straightforward: create a new String object, give it the value Java , and refer x to it. What happens next The VM creates a second String object with the value Java Rules! but nothing refers to it!!! The second String object is instantly lost; no one can ever get to it. The reference variable x still refers to the original String with the value Java . Figure 6-2 shows creating a String object without assigning to a reference. Let s expand this current example. We started with
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String x = "Java"; x.concat(" Rules!"); System.out.println("x = " + x);
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// the output is:
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x = Java
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FIGURE 6-2
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A String object is abandoned upon creation
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Using the String Class (Exam Objective 8.2)
Now let s add
x.toUpperCase(); System.out.println("x = " + x); // the output is still: x = Java
(We actually did just create a new String object with the value JAVA RULES! , but it was lost, and x still refers to the original, unchanged String Java .) How about adding
x.replace('a', 'X'); System.out.println("x = " + x); // the output is still: x = Java
Can you determine what happened The VM created yet another new String object, with the value JXvX , (replacing the a s with X s), but once again this new String was lost, leaving x to refer to the original unchanged and unchangeable String object, with the value Java . In all of these cases we called various String methods to create a new String by altering an existing String, but we never assigned the newly created String to a reference variable. But we can put a small spin on the previous example:
String x = "Java"; x = x.concat(" Rules!"); // Now we're assigning x to the new String System.out.println("x = " + x); // the output will be: // x = Java Rules!
This time, when the VM runs the second line, a new String object is created with the value of Java Rules! , and x is set to reference it. But wait, there s more now the original String object, Java , has been lost, and no one is referring to it. So in both examples we created two String objects and only one reference variable, so one of the two String objects was left out in the cold. See Figure 6-3 for a graphic depiction of this sad story. The dashed line indicates a deleted reference. Let s take this example a little further:
String x = "Java"; x = x.concat(" Rules!"); System.out.println("x = " + x); x.toLowerCase(); //
the output is:
x = Java Rules!
no assignment, create a new, abandoned String // // no assignment, the output is still: x = Java Rules!x =
System.out.println("x = " + x);
6: Java.lang The Math Class, Strings, and Wrappers
x.toLowerCase(); System.out.println("x = " + x);
create a new String, assigned to x // the assignment causes the output: // x = java rules!
The previous discussion contains the keys to understanding Java String immutability. If you really, really get the examples and diagrams, backwards and forwards, you should get 80 percent of the String questions on the exam correct. We will cover more details about Strings next, but make no mistake in terms of bang for your buck, what we ve already covered is by far the most important part of understanding how String objects work in Java. We ll finish this section by presenting an example of the kind of devilish String question you might expect to see on the exam. Take the time to work it out on paper (as a hint, try to keep track of how many objects and reference variables there are, and which ones refer to which).
FIGURE 6-3
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