number of strings created in concatenation operation.

[prateek sharmaa]

hi all,
i have a doubt in string concatenation.

1) how many strings will be created when below line is executed?

String s = new String("abc");

will the system create a string literal "abc" and then a string object on literal "abc"? so will system create 2 strings?

2)how many Strings will be created when below line is executed?

String s1 = "abc";
String s2 = new String("abc");
String s3 = "abc";
String s4 = s1 + s2 + s3;

it will be good if anyone can give little explanation on how the strings will be created.
thanks.

regards,
Pratz

[Stephan van Hulst]

String literals aren't created when the line that references them is being executed. String literals are created when a class is loaded.

Your first example will just create one new String. Your second example will create a new String for s2, and a new String for each concatenation operation, so that's 3 new objects in total. Note that one of those (the String created by s1 + s2) will become eligible for garbage collection after the expression is finished, because there are no references to it anymore.

[prateek sharmaa]

thanks Stephan, that was very much explanatory.

i have one more question that you said String literals are created at the time of class loading so shouldn't we count it as an object?

thanks,
Pratz

[Campbell Ritchie]

Yes, the String literal "abc" is an object in its own right. But you said

when below line is executed

The String literal is loaded as an object into the heap/String literal pool before that line is executed.

[Campbell Ritchie]

prateek sharmaa wrote: . . .

2)how many Strings will be created when below line is executed?

String s1 = "abc";
String s2 = new String("abc");
String s3 = "abc";
String s4 = s1 + s2 + s3;

. . .

Which line? Do you mean all four lines together? I shall let you work it out. I shall also refer you to this very useful JavaRanch Journal article.

[prateek sharmaa]

ok, my mistake
so now can i say that there will be 2 objects in memory after execution of below line?

String s = new String("abc");

one is String literal and other is new String object. right?

[Stephan van Hulst]

Yes, at least those two objects will be in memory, of which the object created from the String literal will be stored in the String pool on the heap.

[prateek sharmaa]

thanks a lot guys. now i am clear about how string objects are created

thanks,
Pratz

[Joanne Neal]

Stephan van Hulst wrote:... and a new String for each concatenation operation, so that's 3 new objects in total. Note that one of those (the String created by s1 + s2) will become eligible for garbage collection after the expression is finished, because there are no references to it anymore.

You're right that the concatenation will create two objects but they won't both be Strings. One will be a StringBuilder. The compiler is clever enough to optimise that bit of code into

String s4 = new StringBuilder(s1).append(s2).append(s3).toString();

In fact the compiler has the option to optimise it in any way it likes, so future versions of Java may optimise it differently. That's why it's recommended to always write

String s4 = s1 + s2 + s3;

rather than

String s4 = new StringBuilder(s1).append(s2).append(s3).toString();

If a later Java version comes up with a better optimisation, the first bit of code will use that optimisation, but if you're explicitly using a StringBuilder then it won't.

[Stephan van Hulst]

Huh cool. I think I've heard this before, but I totally forgot. Thanks Joanne!

[Winston Gutkowski]

Joanne Neal wrote:In fact the compiler has the option to optimise it in any way it likes, so future versions of Java may optimise it differently. That's why it's recommended to always write

String s4 = s1 + s2 + s3;

Hmmm, "overloaded operator as a layer of indirection"... When will it end?

Winston

[Campbell Ritchie]

You need to know about that optimisation.