{Complexity}

Great!
Helped me out regarding String

String s1 = “abc”;
String s2 = “ab”;
s2 = s2 + “c”;
if (s1 == s2)
System.out.println(“Equal”);
else
System.out.println(“Equal”);

Ho According to you s1 is in String constant pool and s2 which is Sring literal created without using new operator should point to s1.
According to you it should display “EQUAL” , but hey buddy run this program and see the behavior

sorry … in hurry i forgot to write not equal…above program will print not equal…

this above code prints not equal becoz
when we are giving statement s2=s2+”c”
prevoius s2 loose its reference ie it no more points to abc
a new object is created as abc and s2 point to it

it no more points to abc ? When did it point either ? s2 was pointing to “ab” but not “abc”

s1=”java”
s2=”";
s2=s1+s2;
if the above condition is checked it will return false. but how it is possible could any one tell me

String s1=”abs”;
String s2 = s2;
System.out.println(s1==s2);

Output:
True

but

String s1=”abs”;
String s2 = s2;
s2=”abs”;
System.out.println(s1==s2);

Output:
false

note that here s2 and s1 point to same memory location till statment String s2=s1;

but when i assign s2=”abs”;
which has same value as s1 and right now its has same memoru location

then its pointing to different object?
why?

its means that while assign s2=”abs”…. it will make new object in constant pool???

@ Munish Gogna Says:
August 28, 2007 at 6:20 am
hi manish.as u r taking strings inside double quotes so they become constants ..so s2 = s2 + “c” this wont wrk..
or change the class .. make it StringBuilder there u can make modification in the string itself

@——Niks Says:
———May 2, 2008 at 12:44 am
hi nikhil
String s2 = s2;
this line wont compiled
make it String s2 = s1;then evrything will work nd giv result
true
true

sorry
in the above post read nikhil as Niks..

hai guy

look at this code some body asked abt it

String s1 = “abc”;
String s2 = “ab”;
s2 = s2 + “c”;
if (s1 == s2)
System.out.println(”Equal”);
else
System.out.println(”false”);

ok this will show false
because when u are trying to add to string a new object will be created so the principle given on the top will work now, as it will create a object in the heap and the heap object refrence will be given back so the s1 and s2 will not at the same address,so it will show u false but if u want to refer to the “abc” in the string constant pool then you can use the “intern()” method as follows so both will refer to same address
String s1=”abc”;
String s2=”ab”;
s2=s2+”c”;
s2=s2.intern();
if(s1==s2)
{
System.out.println(“true”);
}
else
{
System.out.println(“false”);

}

regards
vawani S pati

in this case also it is giving true

thank u for providing this concept. i have doubt in this concept . i am writing two program differently
program 1
class Test{
public static void main(String args[])
{
int i;
String s1=”Hello”;
String s2=”Hello”;
String s3=new String(“Hello”);
String s4=new String(“Hello”);
i=s1.hashCode();
System.out.println(“String s1->”+i); //365478
i=s2.hashCode();
System.out.println(“String s2->”+i); //365478
i=s3.hashCode();
System.out.println(“String s3->”+i); //365478
i=s4.hashCode();
System.out.println(“String s4->”+i); //365478
}
}

That comment is output. that means all hashcode are same .that mean entire content with respect heap and string constant pool are maintain single hashcode. If similar content is available w.r.t to heap as well as string constant pool it is pointed to single hashcode.

ok

observe this program

class Test{
public static void main(String args[])
{
String s1=”hello”;
String s2=”hello”;
String s3=new String(“hello”);
String s4=new String(“hello”);
System.out.println(“w.r.t to heap & string constant pool”);
if(s1==s3)
System.out.println(“String s1 is equal to s”);
else
System.out.println(“String s1 is not equals to s”); // this line executes
System.out.println(“w.r.t string constant pool”);
if(s2==s1)
System.out.println(“String s1 is equal to s”); // this line executes
else
System.out.println(“String s1 is not equals to s”);
System.out.println(“w.r.t to heap “);
if(s3==s4)
System.out.println(“String s1 is equal to s”);
else
System.out.println(“String s1 is not equals to s”); // this line executes
}
}

Observing this two program i got confused . plz clarify this doubt

String.hashCode() returns a value computed from the characters in the string, not the address of the object.

hey guys why did u call Sring litral to a “thing” that is not an “String literal”!! this post is pretty good!