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Problem in String Class

5 replies on 1 page. Most recent reply: Oct 29, 2002 7:47 AM by Don Hill

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siddhartha

Posts: 15
Nickname: saddysan
Registered: Jul, 2002

Problem in String Class Posted: Sep 1, 2002 11:09 AM
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Hello friends,
here is a small program

public class A
{
public static void main(String args[])
{
String st1 = " String ".trim();
String st2 = "String";
String st3 = "String";
System.out.println(st1 == st2);
System.out.println(st2 == st3);
}
}

Output is:
false
true

can somebody explains why it is false in first case and true in second

saddy


Ana

Posts: 12
Nickname: anika
Registered: Aug, 2002

Re: Problem in String Class Posted: Sep 2, 2002 1:47 AM
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The == operator is used to compare references to objects, but not values.
st2 and st3 reference the same object, "String", but st1 references a different one.

If we use the equals() method:

System.out.println(st1.equals(st2));
System.out.println(st2.equals(st3 ));

the output would be:

true
true

siddhartha

Posts: 15
Nickname: saddysan
Registered: Jul, 2002

Re: Problem in String Class Posted: Sep 2, 2002 4:05 AM
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Well thanks for your response, but my doubt is still not over, the question is why " String ".trim() creates another object as we are not using new operator, hence it should be a Strring literal as well and not a new object.

if you say that trim() method internally creates new object than just replace " String ".trim() by
"String".trim() and then it will create String literal.
My question is why it is giving different results. Is there any standard rule by which we can determine if we are going to have String literal or Object.
I am getting confused plz help

saddy

Kishori Sharan

Posts: 211
Nickname: kishori
Registered: Feb, 2002

Re: Problem in String Class Posted: Sep 2, 2002 12:10 PM
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In case of a String if it is created at runtime, which is happeneing in case of "String".trim(), then it is always a new String. Otherwise, if String content is known at compile time then it is treated as String literal. In your case st1 is a new String because it is computed at runtime because you are calling trim() method and st2 and st3 are String literals and that is why st2 and st3 are equal as compared by ==. Please follow this link and read the section on String literals and hopefully your doubt will be cleared.
http://java.sun.com/docs/books/jls/second_edition/html/lexical.doc.html#101083

Thanks
Kishori

Ramasubbu

Posts: 1
Nickname: ramasubbu
Registered: Oct, 2002

Re: Problem in String Class Posted: Oct 29, 2002 7:17 AM
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Hi,
Whenever you create a String object like this,
String a = "test";
String b = "test";
Internally JVM will create object a by allocating memory and use the same memory for object b. So both references are pointing to the same object. Added to that you are checking is a==b?, true(yes a and b are pointing to the same reference). If you check a.equals(b)? true(yes object a value("test") is equal to object b value)

--Ram

Don Hill

Posts: 70
Nickname: ssswdon
Registered: Jul, 2002

Re: Problem in String Class Posted: Oct 29, 2002 7:47 AM
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Another note for you string people,

If you new-up a string as

String a = new String("a");
String b = new String("a");

these string will be pointing to 2 diff strings. Also remember that string are immutable so when ever you apply a method to a string the original string never changes, a new string is created.


HTH

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