Again the string.......help!!

[Amit Das]

Q15. Assuming the following code compiles correctly and prints true, which of the below mentioned concepts are illustrated.
 
compilation unit 1:
 
package MyPackage;
public class MyClass
{
  public static void main (String args[])
  {
    String str1 = "SCJP";
    System.out.println(str1 == AClass.str);
  }
}
class AClass
{
  static String str = "SCJP";
}
 
compilation unit 2:
 
package MyPackage2;
public class AClass
{
  static String str = "SCJP";
} 
 
(1) String objects that are values of constant expressions share unique instances.
(2) Literal strings within the same class in the same package represent references to same String object.
[B]
(3) Strings computed by constant expressions are resolved at compile time and then treated as if they were constant expressions.[/B]
(4) Strings computed at runtime share the same instances and hence are same.
 
 

The answer given is 1, 2 and 3

for thee 4th option, the reason for why it's wrong is given as :
Option 4 is incorrect because the strings which are computed at runtime are distinct and hence DO NOT share same instances. There is one exception to this rule and that is the strings which are constructed from other strings which are declared as final do share same instances.

can anyone explain with some example whatever is here in bold.

thanx
amit

[Animesh Shrivastava]

Option 4 is incorrect because the strings which are computed at runtime are distinct and hence DO NOT share same instances. There is one exception to this rule and that is the strings which are constructed from other strings which are declared as final do share same instances.

hey Amit,
look at the program below, hope it makes u understand

  
public static void main(String[] args) {
                //case 2
String s1 = "Java";
String s2 = "Ranch";
String s3 = "JavaRanch";
String sSum = s1 + s2;
System.out.println("s3 == sSum " + (s3 == sSum)); // false

// Case 2:
final String s1f = "Java";
final String s2f = "Ranch";
final String s3f = "JavaRanch";
String sSumF = s1f + s2f;
System.out.println("s3f == sSumF " + (s3f == sSumF));//True
}

So ur second case explains u ur doubt. U can see that the Strings declared in the second case are final which means they are going to be resolved during compile time and hence variable "ssumF" results to true. Hence the output.