Java Programming - Java.lang Class - Discussion

"abc", "xyz", x, y and "xyzabc" total 5 objects will be created.

xyz will be lost as x now points to new concatenated string. Hence
5-1 = 4 objects.

There are only 3 objects will be created because.

In First line,

String x = new String("xyz");

"xyz" an object is created in memory which is referenced by x(reference variable).if u get the hasCode() of "Hello" and x both will be same because x is referenced to "xyz"

String y = "abc";

This line create one object "abc" which is referenced by y;

And finally when,
x = x + y;

A new object is created "xyzabc" which is referenced by x.
String class is an Immutable string so it is not changed a new object is created when concatenation is performed and is referenced by x.

Please summarize somebody, can we fix the answer as 3 objects.

Please conclude the discussion with proper explanation.

Thanks.

Can anyone show the proof that how 4 objects are getting created. I mean is there any way to show to someone on the machine by executing the program.

Can anyone tell me that why line one creates two object. It should create only one object and this answer is wrong.

I believe "xyz" is a string literal and not string object. So line creates only one object. Also, as stated, literal "xyz" is not lost as its sitting in the stringpool. "xyz" is not a string object.

Line 2 is also creating another string literal in string pool and its not string object.

Line 3 (concatenation) creates another string object.

So in all - 2 objects - one at line 1 and another at line 3.

I agree that Line 1 creates 2 object, and 2nd line 1 object but how 4th object is created?

@Anandi.

String is immutable so new object xyzabc is created.

Proof is:

String x = "xyz";

x == "xyz" false.

Because x referring ingredients of xyz but not to xyz.

And "xyz" is another.

1 line>2.
2> line 1.
3>line 1.

String x = new String ("xyz"); /* one object created here "xyz", note: x is a reference variable only */.

String y = "abc"; /* one object created here "abc", note: y is a reference variable only */.

X = x + y; /* one object created here "xyzabc" referenced by existing x variable */.

So, total 3 objects are only created. Do not get confuse between Objects and reference variables.