RE: Learning Java

Joe,
my background is RPG as well, and I too found the String-Pool process
confusing. What has helped me the most is seeing if the "new" keyword was
used when the variables were defined. If either or both variables are
defined with "new", then the "==" test will be false for the reason you
stated in point#1. 

As far as I know(CMA), the string pool is always in effect, regardless of
where the string is created, and only works for string objects.  If your
compare can "see" both variables, and neither variable was defined with
"new", then the "==" test will work.   For a test, create a class and
subclass it.  Put a static String var in each class, then try comparing them
in your main method.

Here's an example:

class Test {
        static String s2 = "Test";
}

public class TestThis extends Test {
        static String s1 = "Test";
        public static void main(String[] args) {
                if(s1.equals(s2)) {
                        System.out.println("String equals");
                }

                if(s1 == s2) {
                        System.out.println("String == ");
                }
        }
}

Hope this helps,
Mike


-----Original Message-----
From: Joe Teff [mailto:joeteff@earthlink.net]
Sent: Saturday, April 15, 2000 10:47 AM
To: JAVA400-L@midrange.com
Subject: learning Java


I have been spending a lot of time lately learning Java, but have run
across a couple of things I don't understand. My background is RPG
and I don't know C or C++.

1. String x = "100"; String y = "100"; if (x == y) {};
     In this case I thought x would not equal y since they refer to
different
     objects and different memory locations. The book says that they do
     match because the compiler re-uses the same String object if it sees
     the contents match. Is this true only for String objects or other
objects
     treated the same way? Is this true for only String objects in the same
     class or does this optimization occur across classes?

 2. String x = "abc"; String y = "abc"; x  += "def";
     I assume that after the first two statements, both x and y point to the
     same memory location. After the third statement, there are actually
     two objects with different memory locations (x being "abcdef" and y
     being "abc").

Joe Teff

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