String objects

[ashok devaram]

I have a doubt on the number of string objects created in the following code. I got this from Kathy's programmer cert book.
The question is how many string objects are created in the following code

String s1 = "spring";
String s2 = s1 + "summer";
s1.concat("fall");
s2.concat(s1);
s1 += "winter";
System.out.println(s1 + " " + s2);

plz i need some detailed explanation!

thanks in advance

[Mausam M Kakkad]

Hi Ashok,

I think 8 Object will get created.

four objects for literals --> "Spring","Summer","winter","fall"
four other objects for concat operation -->
1) s2 = s1 + "summer"
2) s1.concat(...)
3) s2.concat(...)
4) s1 += ...

I'm having a little doubt here.... while we print by S.o.p....does it create new String object or not...
if it creates then its 9 else 8

someone Please Clarify..

Regards,
Mausam

[Krishna Latha Grandhi]

String s1 = "spring"; // a reference is created in string pool
String s2 = s1 + "summer"; // a reference is created in string poool and one object is created in heap
s1.concat("fall"); // one object is created and it is available for Garbage collection
s2.concat(s1); // one object is created and it is available for Garbage collection
s1 += "winter"; // one object is created in heap
System.out.println(s1 + " " + s2); // one object is created in heap

[Krishna Latha Grandhi]

Sorry , My above post is wrong.

[Mausam M Kakkad]

Thanks Krishna,

That post solved my problem. So Ashok, there will be 9 objects total.
If you draw "reference <---> object diagram u'll come to know with keeping one thing in mind that "String in JAVA is immutable.

Regards,
Mausam

[Krishna Latha Grandhi]

Hi Mausam,

Yeh you are right.

Regards,
krishna.

[Krishna Latha Grandhi]

Hi friends,

I had doubt regarding this question

when using System.out.println

here we also using a literal " ".

I think this adds another object which has the reference in string pool.

In that case the total objects are 10.

plz Any one clarify this doubt..

[Mahendar Reddy]

Yes u r right Hari..
10 objects are created there...

[Mausam M Kakkad]

Oh Yes Hari,

I had forgotten that point there...will take care about that...

Anyways...hope that Ashok has got his doubt cleared..

Regards,
Mausam

[ashok devaram]

hai mahi,kkkad and all

String s1=new String("ashok");//1
String s2="reddy"//2

how many string objects are created at line 1?
how many string objects are created at line 2?
if created where they r created?

[Mausam M Kakkad]

Hi Ashok,

In the first line 2 objects will get created. In the second line 1 object will get created.

The explanation is something like this,
When you specify a String as Literal, one object will get created in the heap and the reference will be put in the String pool. Then when you create a String object by "New" operator, specifically a new object will get created with the value provided in the argument.

String str = new String("mausam");

first "mausam" will get created and its reference will be placed in String literal pool when the class loads. And because of the "New" operator, a new object will get created and the reference will be given to "str".

in case of second statement, we dont have "New" operator. So, just reference of the String will be given.

Hope it will clarify your doubt, otherwise there is a good "Journal" post in JavaRanch itself, which will help you to clarify the concept.

Regards,
Mausam

[ashok devaram]

Thank you very much kakkad.I got the point!

[bama sabapathi]

how come 10 objects?...

String s1 = "spring";
String s2 = s1 + "summer";
s1.concat("fall");
s2.concat(s1);
s1 += "winter";
System.out.println(s1 + " " + s2);

please explain line by line...
Thanks,
Bama.

[Steve Morrow]

String s1 = "spring";      // 1 object, referred to by the literal "spring"
String s2 = s1 + "summer"; // 2 objects; one referred to by the literal"summer",
                           //   one from the concatenation ("springsummer")
s1.concat("fall");         // 2 objects; one referred to by the literal "fall",
                           //   one from the concatenation ("springfall") 
                           //     - (this last reference is thrown away)
s2.concat(s1);             // 1 object from the concatenation ("springsummerspring");
                           //   (this reference is thrown away)
s1 += "winter";            // 2 objects; one referred to by the literal "winter",
                           //   one from the concatenation ("springwinter")
System.out.println(s1 + " " + s2);  
                           // 2 objects; one referred to by " ",
                           //     one from the concatenation ("springwinter springsummer")

Hope this helps.

[bama sabapathi]

Thank you Steve.

-Bama

[achayya matta]

Thanh u very much steve, this time i got it write!

[Mani vannan]

People,
Objects are created in pool or heap. But, their references are created in stack. Am I right?

[Akhilesh Trivedi]

You are right mani. Refer secition on "Local Variables" in K&B's chapter 2.