pooling

[pallavi utukuri]

String a=new String("amit");
String b="arit";
String c=a.replace('m','r');
System.out.println(b==c);
when c String is created replace() returns "arit" which allready exists in the memory pool referenced by b. so why doesnt c refer to b? instead it creats a new String which results b==c to false
Strings always check the memory pool if the new String already
exists in memory, so y doesnt c pick b from pool rather it creates a new string why does this happen
Thanks,
pallavi

[Nathaniel Stoddard]

Pallavi,
Just because two strings have the same value doesn't mean they will be the same object. Take this for example:

public static void main(String[] args) {
    String one = new String("hello");
    String two = new String("hello");
    assert one == two : "See, they're really different!";
}

[pallavi utukuri]

yes ofcourse here they refer to new objects as new is used...... the issue was about pooling ......i got my answer( i think!)
String a=new String("amit");
String c=a.replace('r','m');
a==c now this is true as c is referring to the same pool a does
String b="arit";
String c=b.replace('m','r');
b==c this is also true as c is referrinf to same pool b does
String b="arit";
String c=a.replace('m','r');//c does not refer to b pool as we r using a.replace()
System.out.println(b==c); // therefore its false
i hope my logic is ryt

[Kaustubh Patil]

Use String.intern() to get the ref. fom the same string from the pool. Othrer things cannot assure the same string from the pool.. In fact new String() creates a new string itself irrespective of wheather it exists..
The declaration String s ="aaa";
String s1 = "aaa";
return u the same ref. as its a literal.
try this out.

[Nathaniel Stoddard]

Pallavi,
String.replace will actually create a new String with the "new" operator for you. It doesn't call intern() like Kaustubh mentioned. You would have to call replace().intern() before you could compare it to the other and get a true value (assuming the other was done the same way).

[Yogesh Chhawasaria]

Consider this
When you say

String a = new String("amit");

Two objects are created.

String a = new String("amit");
       |    |             |
       |    |             |
       V    V             |____________|--------------|
    |---------|                        |     amit     |
    |  amit   |                        |--------------|
    |---------|                      Pool of literal Strings
     (Object Space

And foll creates one object
String b = "arit";

String b = "arit";
       |    |             
       |    |             
       |    |__________________________|--------------|
       |                               |     amit     |
       |------------------------------>|--------------|
                                     Pool of literal Strings
     (Object Space

Hence the entire view is
       [a]            [B]
        |              |
        |              |
        |              |               |--------------|
    |---V-----|        |               |     amit     |
    |  amit   |        |               |--------------|
    |---------|        |-------------->|     arit     |  
   (Object Space)                      |--------------|
                                      Pool of literal Strings

Now when u execute
String c = a.replace('m','r'); one new object is created in Object Space

String c = a.replace('m','r');
       |    |             |
       |    |             |
       V    V             |____________|---------------------|
    |---------|                        |arit(already present)|
    |  arit   |                        |---------------------|
    |---------|                      Pool of literal Strings
     (Object Space

So final view is

       [a]      [c]    [b]
        |        |     |
        |        |     |
        |        |     |               |--------------|
    |---V-----|  |     |               |     amit     |
    |  amit   |  |     |               |--------------|
    |---------|  |     |-------------->|     arit     |  
    |  arit   <--|                     |--------------|
    |---------|                        Pool of literal Strings
    (Object Space)

b and c point to diff objs hence false.

[Nathaniel Stoddard]

Yogesh!
That was about the most beautiful reply I've ever seen!! How did you do those cool little diagrams?
[ April 17, 2004: Message edited by: Nathaniel Stoddard ]

[atiqur rahman]

Nice reply
Yogesh Chhawasaria

[Yogesh Chhawasaria]

Thanks !!
I mostly work with diagrams since they seem easy to remember i guess im a visual learner.The pictures make it interesting and it just sticks in mind.
Hey Nathaniel How come with all SCXXY you stil looking for a job. I guess with such suffixes to your name company HR's gotta queuing down ya house.

[pallavi utukuri]

Thanks yogesh and thanks to everyone!