What the output related to string

[Srinivasan Jayakumar]

Hello Friends,
i have preparing for SCJP 5.0, i had a doubt the following question
Given:
11. public String makinStrings() {
12. String s = “Fred”;
13. s = s + “47”;
14. s = s.substring(2, 5);
15. s = s.toUpperCase();
16. return s.toString();
17. }
How many String objects will be created when this method is invoked?
A. 1
B. 2
C. 3
D. 4
E. 5
F. 6

[Punit Singh]

3 objects, search this question, it has been asked lots of time.

[nancy gurudatta]

5

[James Tharakan]

I thought there are only 2 objects created.i.e at line 14 and 15.
Where is the third object :?:

[Punit Singh]

see this link

[Srinivasan Jayakumar]

Hi nancy gurudatta,
how do you say 5 objects can you explain it

[nancy gurudatta]

As far as i know Strings created will be:
"Fred",
"47",
"Fred47"
"ed4"
"ED4"

Please correct me if wrong

[Abhi vijay]

I thought there are only 2 objects created.i.e at line 14 and 15.
Where is the third object

James, what about the object created on LINE13?

[Punit Singh]

12. String s = “Fred”; 
13. s = s + “47”; 

yes line 13 will be created as runtime not at compile time, as String s is not compile time constant.
If it was final String s="Fred", then it would have been created at compile time.

For ex run these codes:

  final String s="abc";//created at compile time
		String s1=s+"def";//created at compile time as it is constant expression
		String s2=s+"def";//created at compile time as it is constant expression
		System.out.println(s1==s2);//true  

  String s="abc";//created at compile time
		String s1=s+"def";//created at runtime so new will be used
		String s2=s+"def";//created at runtime so new will be used
		System.out.println(s1==s2);//false   

[Punit Singh]

Nancy you are wrong, as question is asking about objects created only when this method is called, not total objects, so deduct your compile time objects, count only runtime objects.

[nancy gurudatta]

final String s="abc";//created at compile time
String s1=s+"def";//created at compile time as it is constant expression
String s2=s+"def";//created at compile time as it is constant expression
System.out.println(s1==s2);//true

Hi Punit,
Tell Me..
There will be 2 objects created in above code..
Am I right???

[Punit Singh]

   11. public String makinStrings() { 
12. String s = “Fred”; 
13. s = s + “47”; 
14. s = s.substring(2, 5); 
15. s = s.toUpperCase(); 
16. return s.toString(); 
17. } 

In this code 3 objects will be called as
line 12 is : String s="Fred"; not final String s="Fred"; so it is not a compile time constant that's why
line 13 s=s+"47"; will not be created at compile time, it will be created at run time, when method will be invoked using new operator.

[Punit Singh]

nancy gurudatta wrote:final String s="abc";//created at compile time
String s1=s+"def";//created at compile time as it is constant expression
String s2=s+"def";//created at compile time as it is constant expression
System.out.println(s1==s2);//true

Hi Punit,
Tell Me..
There will be 2 objects created in above code..
Am I right???

Ya you are right here, 2 objects will be created here.

[Michael Tolenruc]

Srinivasan Jayakumar wrote:Hello Friends,
i have preparing for SCJP 5.0, i had a doubt the following question
Given:
11. public String makinStrings() {
12. String s = “Fred”;
13. s = s + “47”;
14. s = s.substring(2, 5);
15. s = s.toUpperCase();
16. return s.toString();
17. }
How many String objects will be created when this method is invoked?
A. 1
B. 2
C. 3
D. 4
E. 5
F. 6

So if we're going to change the question to "How many String Objects will be created?" will be the answer is 5?

[Punit Singh]

yes.

[Jesper de Jong]

Welcome to JavaRanch.

When you post a question copied from a book, mock exam or other source, you are required to quote your sources. This question has been asked many times before, but next time please mention where you copied the question from.

[Abhi vijay]

Can anyone explain about String objects created at runtime and compile time? I am really getting confused.


Here,
String s3="hel"+"lo"; // is it a runtime/ compile time object?

[Punit Singh]

Abhi vijay wrote:Can anyone explain about String objects created at runtime and compile time? I am really getting confused.


Here,
String s3="hel"+"lo"; // is it a runtime/ compile time object?

It will be created at compile time Abhi, as it contains "hel" and "lo" both are constant String literal.

If you says

String s="Abhi";
String s3="hel"+"lo "+s;
Then "hello Abhi" will not be loaded at compile time as we are using String s that is not a compile time constant.

But if you say:
final String s="Abhi";
String s3="hel"+"lo "+s;
Then "hello Abhi" will not be loaded at compile time as we are using final String s that is here now a compile time constant.

[Abhi vijay]

Punit,

  String s ="hello";  // created at compile time.

	String s1 = "hel"; // created at compile time.

	String s2 = "lo"; // created at compile time.

	String s3="hel"+"lo"; // created at compile time as hel and lo both are created at compile time.

	String s4=s1+s2; // created at runtime  LINE1 
  

Here in Line1 s4 should be created at compile time, as s1, s2 both are created at compile time???

[James Tharakan]

It will be created at compile time Abhi, as it contains "hel" and "lo" both are constant String literal.

I thought hel and lo are compile time literals (literal pool).
And during runtime s3 gets a new object hello :?:

[Punit Singh]

James Tharakan wrote:

It will be created at compile time Abhi, as it contains "hel" and "lo" both are constant String literal.

I thought hel and lo are compile time literals (literal pool).
And during runtime s3 gets a new object hello :?:

No s3 will not get new hello, hello will be assigned to it at compile time only. Try this code

  String s5="hel"+"lo";
	System.out.println(s5==s5.intern());  

If both s5 and s5.intern(), referring to same Object than it will return true otherwise false.

[Punit Singh]

Abhi vijay wrote:Punit,

  String s ="hello";  // created at compile time.

	String s1 = "hel"; // created at compile time.

	String s2 = "lo"; // created at compile time.

	String s3="hel"+"lo"; // created at compile time as hel and lo both are created at compile time.

	String s4=s1+s2; // created at runtime  LINE1 
  

Here in Line1 s4 should be created at compile time, as s1, s2 both are created at compile time???

No abhi s4 will not created at compile time, the reason is that s1 and s2 are not declared final, so they will not be treated as compile time constant.

  System.out.println(s4==s4.intern());  

Run this code if it returns true than s1+s2 executed at compile time and s4 assigned reference from the pool.
If false then s4 is assigned new reference at runtime.

Now run this code. And see the output.

final String s1 = "hel"; // created at compile time.
	final String s2 = "lo"; // created at compile time.
	String s3="hel"+"lo"; // created at compile time as hel and lo both are created at compile time.
	String s4=s1+s2;//compile time
	System.out.println(s4==s4.intern());    

Now tell me what you understand from the codes.

[James Tharakan]

EDITED

[Punit Singh]

  String s1 = "hel"; 
	String s2 = "lo"; 
	String s3="hel"+"lo";
	String s4=s1+s2;
	System.out.println(s3==s3.intern());
	System.out.println(s4==s4.intern());  

Now run this code, and tell me what's happening here and why?

[James Tharakan]

Yeah now i think i was wrong .
Hope i wont run into such a confusion again.
And i have erased my previous post to avoid others getting confused.

I wish String class is not a final class. All this confusion could be avoided.

[Punit Singh]

James Tharakan wrote:
I wish String class is not a final class. All this confusion could be avoided.

I did not get meaning of this statement.

[James Tharakan]

hmm..Nothing much ,carry on with your good work..

[Punit Singh]

James did you run this one ?

   
final String s1 = "hel"; // created at compile time.  
    final String s2 = "lo"; // created at compile time.  
    String s3="hel"+"lo"; // created at compile time as hel and lo both are created at compile time.  
    String s4=s1+s2;//compile time  
    System.out.println(s4==s4.intern());  

[James Tharakan]

Yeah , that is clear because of the final modifier,which is compile time constant,so s4 is also done at compile time.

[Punit Singh]

Great