problem with assigning Strings to variables

[kevinn lee]

 class TestString{
public static void main(String args[]){
String s1="abc";
String s2="ab";
String s3=s2+"c";//1
String s4="abc";
System.out.println(s1==s3);//2
System.out.println(s1==s4);
}
}   

why doesnt line 2 give "true"?I mean after line 1 doesn't s3 refer to the same string literal "abc" in String constant pool that s1 is refering to?
Or is there a different mechanism how a new String is created when "+" operator is used.Please explain.
thanks

[Ravinderjit Singh]

Because

String s3=s2+"c";//1  

will create a new String object, means s1 and s3 are referring to two different objects.

[Ravinderjit Singh]

This Link might help you.

[Sandra Bachan]

Hello,

I was trying out your code because I'm puzzled myself.

When I ran your code and the output is

false
true

However, I changed the line 2 to System.out.println(s1.equals(s3)):

class TestString{
	public static void main(String args[]){
		String s1="abc";
		String s2="ab";
		String s3=s2+"c";//1
		String s4="abc";
		System.out.println(s1.equals(s3));//2
		System.out.println(s1==s4);
	}
}  

Now the output is

true
true

On the other hand, I'm not sure why s1==s4 outputs to true; my understanding is that String s1, s2, s3 and s4 refer to different objects on the heap - then again, I haven't read the chapter on Strings yet...

[kevinn lee]

Ravinderjit Singh wrote:Because

String s3=s2+"c";//1  

will create a new String object, means s1 and s3 are referring to two different objects.

you mean

String s3=s2+"c";

is like

    s3=new String("abc"); 

rather than

s3="abc";    

is it?
thanks

[Jim Hoglund]

The string pool mechanism is at work here. To conserve memory, the compiler will
at times, share a string object if sX.equals(sY). Apparently in your case it put "abc" in
the pool and used it again. (Maybe for s1, s3 and s4).)

Jim ... ...

[kevinn lee]

Sandra Bachan wrote:Hello,

hello
go to the link given by Ravinderjit
it explains what is special about Strings
thanks for replying

[kevinn lee]

Jim Hoglund wrote:The string pool mechanism is at work here. To conserve memory, the compiler will
at times, share a string object if sX.equals(sY). Apparently in your case it put "abc" in
the pool and used it again. (Maybe for s1, s3 and s4).)

Jim ... ...

but it seems as if "abc" is not shared here since "s1==s3" turns out to be false.
thanks

[Sandra Bachan]

but it seems as if "abc" is not shared here since "s1==s3" turns out to be false.

From reading the link, my understanding is that if a string is concatenated, it will create a new object. However, if you apply intern(), then it looks for pre-existing String object in memory for s3 to point to. The pre-existing object that s3 originally pointed to is now eligible for garbage collection:

class TestString{
	
	public static void main(String args[]){

		String s1="abc";

		String s2="ab";

		String s3=s2+"c";//1

		String s4="abc";

		System.out.println(s1==s3);//2
		
		System.out.println(s1==s4);
		
		s3 = s3.intern();
		
		System.out.println("After Intern Function on s3:");
		
		System.out.println(s1==s3);//2



	}
}

Output:

false
true
After Intern Function on s3:
true

[kevinn lee]

thanks a lot sandra