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Describe the significance of the immutability of String objects. This section covers the String and StringBuffer classes. The key concepts we ll cover will help you understand that once a String object is created, it can never be changed so what is happening when a String object seems to be changing We ll find out. We ll also cover the differences between the String and StringBuffer classes and when to use which.
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Strings Are Immutable Objects
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Let s start with a little background information about strings. Strictly speaking you may not need this information for the test, but a little context will help you learn what you do have to know. Handling strings of characters is a fundamental aspect of most programming languages. In Java, each character in a string is a 16-bit
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Unicode character. Because Unicode characters are 16 bits (not the skimpy 7 or 8 bits that ASCII provides), a rich, international set of characters is easily represented in Unicode. In Java, strings are objects. Just like other objects, you can create an instance of a String with the new keyword, as follows:
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String s = new String();
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This line of code creates a new object of class String, and assigns the reference variable s to it. So far String objects seem just like other objects. Now, let s give the String a value:
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s = "abcdef";
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As you might expect the String class has about a zillion constructors, so you can use a more efficient shortcut:
String s = new String("abcdef");
And just because you ll use strings all the time, you can even say this:
String s = "abcdef";
There are some subtle differences between these options that we ll discuss later, but what they have in common is that they all create a new String object, with a value of abcdef , and assign it to a reference variable s. Now let s say that you want a second reference to the String object referred to by s:
String s2 = s; // refer s2 to the same String as s
So far so good. String objects seem to be behaving just like other objects, so what s all the fuss about The certification objective states: describe the significance of the immutability of String objects. Ah-ha! Immutability! (What the heck is immutability ) Once you have assigned a String a value, that value can never change it s immutable, frozen solid, won t budge, fini, done. (We ll also talk about why later, don t let us forget.) The good news is that while the String object is immutable, its reference variable is not, so to continue with our previous example:
s = s.concat(" more stuff"); // the concat() method 'appends // a literal to the end
6: Java.lang The Math Class, Strings, and Wrappers
Now wait just a minute, didn t we just say that Strings were immutable So what s all this appending to the end of the string talk Excellent question; let s look at what really happened The VM took the value of String s (which was abcdef ), and tacked more stuff onto the end, giving us the value abcdef more stuff . Since Strings are immutable, the VM couldn t stuff this new String into the old String referenced by s, so it created a new String object, gave it the value abcdef more stuff , and made s refer to it. At this point in our example, we have two String objects: the first one we created, with the value abcdef , and the second one with the value abcdef more stuff . Technically there are now three String objects, because the literal argument to concat more stuff is itself a new String object. But we have references only to abcdef (referenced by s2) and abcdef more stuff (referenced by s). What if we didn t have the foresight or luck to create a second reference variable for the abcdef String before we called: s = s.concat( more stuff ); In that case the original, unchanged String containing abcdef would still exist in memory, but it would be considered lost. No code in our program has any way to reference it it is lost to us. Note, however, that the original abcdef String didn t change (it can t, remember, it s immutable); only the reference variable s was changed, so that it would refer to a different String. Figure 6-1 shows what happens on the heap when you reassign a reference variable. Note that the dashed line indicates a deleted reference. To review our first example:
String s = "abcdef"; String s2 = s; // // // // create a new String object, with value "abcdef", refer s to it create a 2nd reference variable referring to the same String // // // // // // create a new String object, with value "abcdef more stuff", refer s to it. (change s's reference from the old String to the new String. ( Remember s2 is still referring to the original "abcdef" String.
s = s.concat(" more stuff");
Using the String Class (Exam Objective 8.2)
FIGURE 6-1
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