string function problem

How many objects of String are created?
string x="xyz";
x=x+"pqr";
tHANKS

Originally posted by priyanshu bhardwaj:
How many objects of String are created?
string x="xyz";
x=x+"pqr";
tHANKS

In this case, totally three strings("xyz", "pqr", "xyzpqr") will be created in the constant pool. To see more on this, click the following link
String constant pool

Originally posted by Periakaruppan Thiagarajan:
in the constant pool.

No, actually -- only the two literal (quoted) strings are in the constant pool. The concatenation is done at runtime.

I'm going to move this to SCJP, just to make the point that these sorts of questions belong there, not here.

SIR
iCAN'T UNDERSTAND WHAT IS THE RIGHT ANSWER ONE OR TWO OR THREE?
Thanks

according to my understanding when u create String x="xyz"..then any matches to this literal "xyz" in the String pool is found out..if any match is found...then it is reffered by variable x...therefore here new object is not created....when we performed the concatenation in the later step....new object is created and is reffered back to variable x....hence only one object is created...this is only a guess...waiting for the correct answer...... wats the ans...

Hi,

Only 2 instances will be created since the Operator "New" isn't used while creating the string.

Cheers

How many objects of String are created?
string x="xyz";
x=x+"pqr";
tHANKS

Since Strings are immutable , Of total three Obj's are created.
1 -> "xyz" ; even though it exists in memory, it is considered lost after reassigning to 'x'.
2->

How many objects of String are created?
string x="xyz";
x=x+"pqr";
tHANKS

Since Strings are immutable , Of total three Obj's are created.
1 -> "xyz" ; even though it exists in memory, it is considered lost after reassigning to 'x'.
2-> "xyzpqr"
3-> "pqr" itself is one object.
Plz suggest, if I'm wrong.

regards,
Sudhir

My answer is 3

Hi

Three objects created.

I go with Ernest, only 2 objects are created,

String literals are created at compile time and not at runtime

string x="xyz";
x=x+"pqr";

only "xyz" and "pqr" are created and not "xyzpqr"

there was a previous discussion already in this forum, plse find the link below

http://www.coderanch.com/t/246349/java-programmer-SCJP/certification/Journal-Article-SCJP-Tip-Line

Hi All,
Please confirm the correct answer is two or three, i'm kind of cofused.
Regards,
Simi

i think 3 objects will be created 2 in pool and one not in pool
x="xyz"; creates a string object "xyz" in pool
x=x+"pqr";it creates a string object "pqr" in pool and since value of variable x is evaluated at runtime the new string object "xyzpqr" will be created outside of pool

this question was discussed just a few days ago, after many thoughts i am pretty sure that three objects are created.

my references are these three:

http://www.javaranch.com/journal/200409/ScjpTipLine-StringsLiterally.html, where the situation String s = new ("someString"); is explained explicitly

the java language specification. quote: "Each string literal is a reference to an instance of class String",

K & B book page 360

anyway, if somebody has a strong counter-argument, then i am able to resign from this position

greetings

Hi All,

The below example from K&B should help.

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

The result of this code fragment is �spring winter spring summer�.
There are two reference variables, s1 and s2. There were a total of eight String objects created as follows: �spring�, �summer � (lost), �spring summer�, �fall� (lost), �spring fall� (lost), �spring summer spring� (lost), �winter� (lost), �spring winter� (at this point �spring� is lost). Only two of the eight String objects are not lost in this process.

See String Objects

[ January 30, 2006: Message edited by: Lalitha Vydyula ]
[ January 30, 2006: Message edited by: Lalitha Vydyula ]