identifier

[Jian Yi]

public class SubstringTest
{
   public static void main(String args[]) 
   {
      String String = "string isa string";
      System.out.println(String.substring(3, 6));
   }
}

Why does this code even compile? Isn't String a keyword? Isn't it true that a keyword can't be used as an identifier?
thanks,
Jenny

[Corey McGlone]

Originally posted by Jenny Yin:
Isn't String a keyword? Isn't it true that a keyword can't be used as an identifier?
thanks,
Jenny

String is not a keyword - it's the name of a class. Class names are still legal identifiers in Java. It is true, however, that keywords can't be used as identifiers. For example, you can't use "this" as an identifier.
Corey
[ April 24, 2002: Message edited by: Corey McGlone ]

[Junilu Lacar]

Jenny,
In the code you gave, the local String variable effectively shadows the java.lang.String class. Consider the following code:

  
 1:  public class Q1 {
 2:      public static void main (String[] args) {
 3:          StringBuffer String = new StringBuffer("test");
 4:          String y = "foo";  
 5:          int x = 5;
 6:          System.out.println(y + String);
 7:          System.out.println(java.lang.String.valueOf(x)); 
 8:      }
 9:  }

This code compiles and produces the output:
footest
5
On line 7, you need to fully qualify the reference to java.lang.String otherwise you will get a compiler error because the compiler will use the local String variable.
Notice the compiler is smart enough (well, actually the guys who wrote the compiler are the smarties) to use the context to recognize that 'String' on line 4 refers to the shadowed java.lang.String rather than the local variable String.
HTH,
Junilu

[Jian Yi]

Corey,
Thanks for pointing that out.
Junilu,
That's a fun exercise you showed me. Thanks.
-Jenny