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An interview question: What's the number of String objects being created?


Is an array a primitive type or an object (or something else entirely)?How do I create a Java string from the contents of a file?Creating multiline strings in JavaScriptCount the number of occurrences of a character in a string in JavascriptHow do I check if a string contains another string in Objective-C?Convert JS object to JSON stringConverting an object to a stringPythonic way to create a long multi-line stringWhy is String immutable in Java?Where is the new Object of String created when we concat using + operator






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








6















I was asked in an interview about the number of objects that will be created on the given problem:



String str1 = "First";
String str2 = "Second";
String str3 = "Third";
String str4 = str1 + str2 + str3;


So I am not sure that the answer I gave was correct or not.



I answered that there would be 6 objects created in the string pool.



3 would be for all three variables.

1 would be for str1 + str2 (let's say str).

1 would be for str2 + str3.

1 would be for the str + str3 (str = str1 + str2).



Please help me out on this confusion.










share|improve this question









New contributor



bhpsh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • 2





    Why would str2 + str3 be one of the objects?

    – Jacob G.
    8 hours ago






  • 3





    I am not entirely sure whether it is defined in the JLS, but when concatenating Strings, the compiler normally generates a StringBuilder to concatenate the Strings. I am not entirely sure how the StringBuilder internally handles the concatenation, but I would say that at least five Objects are created: one for str1 to str3, one StringBuilder and the final String for String4.

    – Turing85
    8 hours ago











  • Update: The JLS actually defines that "a Java compiler may use the StringBuffer class or a similar technique to reduce the number of intermediate String objects that are created by evaluation of an expression." In other words: without further information, the question cannot be definitively answered.

    – Turing85
    8 hours ago











  • Nothing against you, but I hate questions like that. What is the deeper meaning of such questions? Especially in an interview? What skills does the interviewer want to find out?

    – RedCam
    7 hours ago











  • if the first 3 variables were defined final I would say no object is being created (at that code segment)

    – Carlos Heuberger
    7 hours ago

















6















I was asked in an interview about the number of objects that will be created on the given problem:



String str1 = "First";
String str2 = "Second";
String str3 = "Third";
String str4 = str1 + str2 + str3;


So I am not sure that the answer I gave was correct or not.



I answered that there would be 6 objects created in the string pool.



3 would be for all three variables.

1 would be for str1 + str2 (let's say str).

1 would be for str2 + str3.

1 would be for the str + str3 (str = str1 + str2).



Please help me out on this confusion.










share|improve this question









New contributor



bhpsh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
















  • 2





    Why would str2 + str3 be one of the objects?

    – Jacob G.
    8 hours ago






  • 3





    I am not entirely sure whether it is defined in the JLS, but when concatenating Strings, the compiler normally generates a StringBuilder to concatenate the Strings. I am not entirely sure how the StringBuilder internally handles the concatenation, but I would say that at least five Objects are created: one for str1 to str3, one StringBuilder and the final String for String4.

    – Turing85
    8 hours ago











  • Update: The JLS actually defines that "a Java compiler may use the StringBuffer class or a similar technique to reduce the number of intermediate String objects that are created by evaluation of an expression." In other words: without further information, the question cannot be definitively answered.

    – Turing85
    8 hours ago











  • Nothing against you, but I hate questions like that. What is the deeper meaning of such questions? Especially in an interview? What skills does the interviewer want to find out?

    – RedCam
    7 hours ago











  • if the first 3 variables were defined final I would say no object is being created (at that code segment)

    – Carlos Heuberger
    7 hours ago













6












6








6








I was asked in an interview about the number of objects that will be created on the given problem:



String str1 = "First";
String str2 = "Second";
String str3 = "Third";
String str4 = str1 + str2 + str3;


So I am not sure that the answer I gave was correct or not.



I answered that there would be 6 objects created in the string pool.



3 would be for all three variables.

1 would be for str1 + str2 (let's say str).

1 would be for str2 + str3.

1 would be for the str + str3 (str = str1 + str2).



Please help me out on this confusion.










share|improve this question









New contributor



bhpsh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











I was asked in an interview about the number of objects that will be created on the given problem:



String str1 = "First";
String str2 = "Second";
String str3 = "Third";
String str4 = str1 + str2 + str3;


So I am not sure that the answer I gave was correct or not.



I answered that there would be 6 objects created in the string pool.



3 would be for all three variables.

1 would be for str1 + str2 (let's say str).

1 would be for str2 + str3.

1 would be for the str + str3 (str = str1 + str2).



Please help me out on this confusion.







java string string-concatenation string-pool string-building






share|improve this question









New contributor



bhpsh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.










share|improve this question









New contributor



bhpsh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








share|improve this question




share|improve this question








edited 7 hours ago









Andrew Tobilko

32.9k10 gold badges53 silver badges102 bronze badges




32.9k10 gold badges53 silver badges102 bronze badges






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asked 8 hours ago









bhpshbhpsh

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373 bronze badges




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Check out our Code of Conduct.












  • 2





    Why would str2 + str3 be one of the objects?

    – Jacob G.
    8 hours ago






  • 3





    I am not entirely sure whether it is defined in the JLS, but when concatenating Strings, the compiler normally generates a StringBuilder to concatenate the Strings. I am not entirely sure how the StringBuilder internally handles the concatenation, but I would say that at least five Objects are created: one for str1 to str3, one StringBuilder and the final String for String4.

    – Turing85
    8 hours ago











  • Update: The JLS actually defines that "a Java compiler may use the StringBuffer class or a similar technique to reduce the number of intermediate String objects that are created by evaluation of an expression." In other words: without further information, the question cannot be definitively answered.

    – Turing85
    8 hours ago











  • Nothing against you, but I hate questions like that. What is the deeper meaning of such questions? Especially in an interview? What skills does the interviewer want to find out?

    – RedCam
    7 hours ago











  • if the first 3 variables were defined final I would say no object is being created (at that code segment)

    – Carlos Heuberger
    7 hours ago












  • 2





    Why would str2 + str3 be one of the objects?

    – Jacob G.
    8 hours ago






  • 3





    I am not entirely sure whether it is defined in the JLS, but when concatenating Strings, the compiler normally generates a StringBuilder to concatenate the Strings. I am not entirely sure how the StringBuilder internally handles the concatenation, but I would say that at least five Objects are created: one for str1 to str3, one StringBuilder and the final String for String4.

    – Turing85
    8 hours ago











  • Update: The JLS actually defines that "a Java compiler may use the StringBuffer class or a similar technique to reduce the number of intermediate String objects that are created by evaluation of an expression." In other words: without further information, the question cannot be definitively answered.

    – Turing85
    8 hours ago











  • Nothing against you, but I hate questions like that. What is the deeper meaning of such questions? Especially in an interview? What skills does the interviewer want to find out?

    – RedCam
    7 hours ago











  • if the first 3 variables were defined final I would say no object is being created (at that code segment)

    – Carlos Heuberger
    7 hours ago







2




2





Why would str2 + str3 be one of the objects?

– Jacob G.
8 hours ago





Why would str2 + str3 be one of the objects?

– Jacob G.
8 hours ago




3




3





I am not entirely sure whether it is defined in the JLS, but when concatenating Strings, the compiler normally generates a StringBuilder to concatenate the Strings. I am not entirely sure how the StringBuilder internally handles the concatenation, but I would say that at least five Objects are created: one for str1 to str3, one StringBuilder and the final String for String4.

– Turing85
8 hours ago





I am not entirely sure whether it is defined in the JLS, but when concatenating Strings, the compiler normally generates a StringBuilder to concatenate the Strings. I am not entirely sure how the StringBuilder internally handles the concatenation, but I would say that at least five Objects are created: one for str1 to str3, one StringBuilder and the final String for String4.

– Turing85
8 hours ago













Update: The JLS actually defines that "a Java compiler may use the StringBuffer class or a similar technique to reduce the number of intermediate String objects that are created by evaluation of an expression." In other words: without further information, the question cannot be definitively answered.

– Turing85
8 hours ago





Update: The JLS actually defines that "a Java compiler may use the StringBuffer class or a similar technique to reduce the number of intermediate String objects that are created by evaluation of an expression." In other words: without further information, the question cannot be definitively answered.

– Turing85
8 hours ago













Nothing against you, but I hate questions like that. What is the deeper meaning of such questions? Especially in an interview? What skills does the interviewer want to find out?

– RedCam
7 hours ago





Nothing against you, but I hate questions like that. What is the deeper meaning of such questions? Especially in an interview? What skills does the interviewer want to find out?

– RedCam
7 hours ago













if the first 3 variables were defined final I would say no object is being created (at that code segment)

– Carlos Heuberger
7 hours ago





if the first 3 variables were defined final I would say no object is being created (at that code segment)

– Carlos Heuberger
7 hours ago












5 Answers
5






active

oldest

votes


















9















With the given information, the question cannot be definitely answered. As is stated in the JLS, §15.18.1:




... To increase the performance of repeated string concatenation, a Java compiler may use the StringBuffer class or a similar technique to reduce the number of intermediate String objects that are created by evaluation of an expression.




This means that the answer depends at least on the concrete Java compiler used.



I think the best we can do is give an interval as answer:



  • a smart compiler may be able to infer that str1 to str3 are never used and fold the concatenation during compilation, such that only one String-object is created (the one referenced by str4)

  • The maximum sensible number of Strings created should be 5: one each for str1 to str3, one for tmp = str1 + str2 and one for str4 = tmp + str3.

So... my answer would be "something between one to five String-objects". As to the total number of objects created just for this operation... I do not know. This may also depend how exactly e.g. StringBuffer is implemented.



As an aside: I wonder what the reason behind asking such questions is. Normally, one does not need to care about those details.






share|improve this answer






















  • 3





    "I wonder what the reason behind asking such questions is." - It reminds me of the "I-read-it-yesterday-I-will-ask-it-tomorrow" type of interviewers

    – Andrew Tobilko
    7 hours ago



















3















Java 8 will likely create 5 objects:



  • 3 for the 3 variables

  • 1 StringBuilder

  • 1 for the concatenated String

With Java 9 things changed though and String concatenation does not use StringBuilder anymore.






share|improve this answer




















  • 3





    Good answer, but for Java 8, I think you're overlooking the char[] object used by the implementation of StringBuilder.

    – Andy Thomas
    8 hours ago







  • 1





    stackoverflow.com/questions/12806739/…

    – 17slim
    7 hours ago






  • 4





    @3limin4t0r "char[] is a primitive value" - Nope.

    – Turing85
    7 hours ago







  • 2





    @AndyThomas if you count the char[] you will end up with even more as also String uses char[] and some constructors also create copies of char[].

    – Puce
    7 hours ago











  • That's a good point actually, which leads me to believe @Turing85's answer is more correct without further clarification from the interviewers.

    – 17slim
    7 hours ago


















2















It should be 5:



  • three for the three literals (assigned to str1, str2 and str3)


  • one for str1 + str2


  • one for (result from the previous operation) + str3 (assigned to str4)






share|improve this answer



























  • str1 + str2 will not be concatenated to a String in all Java since quite many years.

    – Puce
    8 hours ago


















2















Any answer to your question will be dependant on the JVM implementation and the Java version currently being used. It's a stupid question to ask in an interview.



Java 1.8.0_201



On my machine, with Java 1.8.0_201, your snippet results in this bytecode



 L0
LINENUMBER 13 L0
LDC "First"
ASTORE 1
L1
LINENUMBER 14 L1
LDC "Second"
ASTORE 2
L2
LINENUMBER 15 L2
LDC "Third"
ASTORE 3
L3
LINENUMBER 16 L3
NEW java/lang/StringBuilder
DUP
INVOKESPECIAL java/lang/StringBuilder.<init> ()V
ALOAD 1
INVOKEVIRTUAL java/lang/StringBuilder.append (Ljava/lang/String;)Ljava/lang/StringBuilder;
ALOAD 2
INVOKEVIRTUAL java/lang/StringBuilder.append (Ljava/lang/String;)Ljava/lang/StringBuilder;
ALOAD 3
INVOKEVIRTUAL java/lang/StringBuilder.append (Ljava/lang/String;)Ljava/lang/StringBuilder;
INVOKEVIRTUAL java/lang/StringBuilder.toString ()Ljava/lang/String;
ASTORE 4
L4
LINENUMBER 19 L4
RETURN


which proves that 5 objects are being created (3 constant String literals, 1 StringBuilder, 1 dynamically produced String instance by StringBuilder#toString).



Java 12.0.2



On my machine, with Java 12.0.2, the bytecode is



 L0
LINENUMBER 13 L0
LDC "First"
ASTORE 1
L1
LINENUMBER 14 L1
LDC "Second"
ASTORE 2
L2
LINENUMBER 15 L2
LDC "Third"
ASTORE 3
L3
LINENUMBER 16 L3
ALOAD 1
ALOAD 2
ALOAD 3
INVOKEDYNAMIC makeConcatWithConstants(Ljava/lang/String;Ljava/lang/String;Ljava/lang/String;)Ljava/lang/String; [
// handle kind 0x6 : INVOKESTATIC
java/lang/invoke/StringConcatFactory.makeConcatWithConstants(Ljava/lang/invoke/MethodHandles$Lookup;Ljava/lang/String;Ljava/lang/invoke/MethodType;Ljava/lang/String;[Ljava/lang/Object;)Ljava/lang/invoke/CallSite;
// arguments:
"u0001u0001u0001"
]
ASTORE 4
L4
LINENUMBER 17 L4
RETURN


which magically changes "the correct answer" to 4 objects since there is no intermediate String[Buffer/Builder] involved.






share|improve this answer






















  • 1





    each string also contains a byte[], same for StringBuilder (and first one will also create a static byte[0]) and eventually (haven't checked) the StringBuilder can create a new, bigger buffer (if needed) (just proving your 2nd sentence true)

    – Carlos Heuberger
    7 hours ago












  • thinking about it, is LDC creating a (new) Object? or just using one? (the literal could have been used before)

    – Carlos Heuberger
    7 hours ago



















0















Concatenation operation doesn't create those many String objects. It creates aStringBuilder and then appends the strings. So there may be 5 objects,
3 (variables) + 1 (sb) + 1 (Concatenated string).






share|improve this answer



























  • It might be 5.

    – Nexevis
    7 hours ago











  • As was stated many times, the JLS says that "a Java compiler may use the StringBuffer class or a similar technique to reduce the number of intermediate String objects that are created by evaluation of an expression." Emphasis is on may.

    – Turing85
    7 hours ago












  • @Turing85 agreed. Updated.

    – javaaddict
    7 hours ago













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5 Answers
5






active

oldest

votes








5 Answers
5






active

oldest

votes









active

oldest

votes






active

oldest

votes









9















With the given information, the question cannot be definitely answered. As is stated in the JLS, §15.18.1:




... To increase the performance of repeated string concatenation, a Java compiler may use the StringBuffer class or a similar technique to reduce the number of intermediate String objects that are created by evaluation of an expression.




This means that the answer depends at least on the concrete Java compiler used.



I think the best we can do is give an interval as answer:



  • a smart compiler may be able to infer that str1 to str3 are never used and fold the concatenation during compilation, such that only one String-object is created (the one referenced by str4)

  • The maximum sensible number of Strings created should be 5: one each for str1 to str3, one for tmp = str1 + str2 and one for str4 = tmp + str3.

So... my answer would be "something between one to five String-objects". As to the total number of objects created just for this operation... I do not know. This may also depend how exactly e.g. StringBuffer is implemented.



As an aside: I wonder what the reason behind asking such questions is. Normally, one does not need to care about those details.






share|improve this answer






















  • 3





    "I wonder what the reason behind asking such questions is." - It reminds me of the "I-read-it-yesterday-I-will-ask-it-tomorrow" type of interviewers

    – Andrew Tobilko
    7 hours ago
















9















With the given information, the question cannot be definitely answered. As is stated in the JLS, §15.18.1:




... To increase the performance of repeated string concatenation, a Java compiler may use the StringBuffer class or a similar technique to reduce the number of intermediate String objects that are created by evaluation of an expression.




This means that the answer depends at least on the concrete Java compiler used.



I think the best we can do is give an interval as answer:



  • a smart compiler may be able to infer that str1 to str3 are never used and fold the concatenation during compilation, such that only one String-object is created (the one referenced by str4)

  • The maximum sensible number of Strings created should be 5: one each for str1 to str3, one for tmp = str1 + str2 and one for str4 = tmp + str3.

So... my answer would be "something between one to five String-objects". As to the total number of objects created just for this operation... I do not know. This may also depend how exactly e.g. StringBuffer is implemented.



As an aside: I wonder what the reason behind asking such questions is. Normally, one does not need to care about those details.






share|improve this answer






















  • 3





    "I wonder what the reason behind asking such questions is." - It reminds me of the "I-read-it-yesterday-I-will-ask-it-tomorrow" type of interviewers

    – Andrew Tobilko
    7 hours ago














9














9










9









With the given information, the question cannot be definitely answered. As is stated in the JLS, §15.18.1:




... To increase the performance of repeated string concatenation, a Java compiler may use the StringBuffer class or a similar technique to reduce the number of intermediate String objects that are created by evaluation of an expression.




This means that the answer depends at least on the concrete Java compiler used.



I think the best we can do is give an interval as answer:



  • a smart compiler may be able to infer that str1 to str3 are never used and fold the concatenation during compilation, such that only one String-object is created (the one referenced by str4)

  • The maximum sensible number of Strings created should be 5: one each for str1 to str3, one for tmp = str1 + str2 and one for str4 = tmp + str3.

So... my answer would be "something between one to five String-objects". As to the total number of objects created just for this operation... I do not know. This may also depend how exactly e.g. StringBuffer is implemented.



As an aside: I wonder what the reason behind asking such questions is. Normally, one does not need to care about those details.






share|improve this answer















With the given information, the question cannot be definitely answered. As is stated in the JLS, §15.18.1:




... To increase the performance of repeated string concatenation, a Java compiler may use the StringBuffer class or a similar technique to reduce the number of intermediate String objects that are created by evaluation of an expression.




This means that the answer depends at least on the concrete Java compiler used.



I think the best we can do is give an interval as answer:



  • a smart compiler may be able to infer that str1 to str3 are never used and fold the concatenation during compilation, such that only one String-object is created (the one referenced by str4)

  • The maximum sensible number of Strings created should be 5: one each for str1 to str3, one for tmp = str1 + str2 and one for str4 = tmp + str3.

So... my answer would be "something between one to five String-objects". As to the total number of objects created just for this operation... I do not know. This may also depend how exactly e.g. StringBuffer is implemented.



As an aside: I wonder what the reason behind asking such questions is. Normally, one does not need to care about those details.







share|improve this answer














share|improve this answer



share|improve this answer








edited 7 hours ago

























answered 8 hours ago









Turing85Turing85

7,1863 gold badges19 silver badges42 bronze badges




7,1863 gold badges19 silver badges42 bronze badges










  • 3





    "I wonder what the reason behind asking such questions is." - It reminds me of the "I-read-it-yesterday-I-will-ask-it-tomorrow" type of interviewers

    – Andrew Tobilko
    7 hours ago













  • 3





    "I wonder what the reason behind asking such questions is." - It reminds me of the "I-read-it-yesterday-I-will-ask-it-tomorrow" type of interviewers

    – Andrew Tobilko
    7 hours ago








3




3





"I wonder what the reason behind asking such questions is." - It reminds me of the "I-read-it-yesterday-I-will-ask-it-tomorrow" type of interviewers

– Andrew Tobilko
7 hours ago






"I wonder what the reason behind asking such questions is." - It reminds me of the "I-read-it-yesterday-I-will-ask-it-tomorrow" type of interviewers

– Andrew Tobilko
7 hours ago














3















Java 8 will likely create 5 objects:



  • 3 for the 3 variables

  • 1 StringBuilder

  • 1 for the concatenated String

With Java 9 things changed though and String concatenation does not use StringBuilder anymore.






share|improve this answer




















  • 3





    Good answer, but for Java 8, I think you're overlooking the char[] object used by the implementation of StringBuilder.

    – Andy Thomas
    8 hours ago







  • 1





    stackoverflow.com/questions/12806739/…

    – 17slim
    7 hours ago






  • 4





    @3limin4t0r "char[] is a primitive value" - Nope.

    – Turing85
    7 hours ago







  • 2





    @AndyThomas if you count the char[] you will end up with even more as also String uses char[] and some constructors also create copies of char[].

    – Puce
    7 hours ago











  • That's a good point actually, which leads me to believe @Turing85's answer is more correct without further clarification from the interviewers.

    – 17slim
    7 hours ago















3















Java 8 will likely create 5 objects:



  • 3 for the 3 variables

  • 1 StringBuilder

  • 1 for the concatenated String

With Java 9 things changed though and String concatenation does not use StringBuilder anymore.






share|improve this answer




















  • 3





    Good answer, but for Java 8, I think you're overlooking the char[] object used by the implementation of StringBuilder.

    – Andy Thomas
    8 hours ago







  • 1





    stackoverflow.com/questions/12806739/…

    – 17slim
    7 hours ago






  • 4





    @3limin4t0r "char[] is a primitive value" - Nope.

    – Turing85
    7 hours ago







  • 2





    @AndyThomas if you count the char[] you will end up with even more as also String uses char[] and some constructors also create copies of char[].

    – Puce
    7 hours ago











  • That's a good point actually, which leads me to believe @Turing85's answer is more correct without further clarification from the interviewers.

    – 17slim
    7 hours ago













3














3










3









Java 8 will likely create 5 objects:



  • 3 for the 3 variables

  • 1 StringBuilder

  • 1 for the concatenated String

With Java 9 things changed though and String concatenation does not use StringBuilder anymore.






share|improve this answer













Java 8 will likely create 5 objects:



  • 3 for the 3 variables

  • 1 StringBuilder

  • 1 for the concatenated String

With Java 9 things changed though and String concatenation does not use StringBuilder anymore.







share|improve this answer












share|improve this answer



share|improve this answer










answered 8 hours ago









PucePuce

29.1k9 gold badges61 silver badges117 bronze badges




29.1k9 gold badges61 silver badges117 bronze badges










  • 3





    Good answer, but for Java 8, I think you're overlooking the char[] object used by the implementation of StringBuilder.

    – Andy Thomas
    8 hours ago







  • 1





    stackoverflow.com/questions/12806739/…

    – 17slim
    7 hours ago






  • 4





    @3limin4t0r "char[] is a primitive value" - Nope.

    – Turing85
    7 hours ago







  • 2





    @AndyThomas if you count the char[] you will end up with even more as also String uses char[] and some constructors also create copies of char[].

    – Puce
    7 hours ago











  • That's a good point actually, which leads me to believe @Turing85's answer is more correct without further clarification from the interviewers.

    – 17slim
    7 hours ago












  • 3





    Good answer, but for Java 8, I think you're overlooking the char[] object used by the implementation of StringBuilder.

    – Andy Thomas
    8 hours ago







  • 1





    stackoverflow.com/questions/12806739/…

    – 17slim
    7 hours ago






  • 4





    @3limin4t0r "char[] is a primitive value" - Nope.

    – Turing85
    7 hours ago







  • 2





    @AndyThomas if you count the char[] you will end up with even more as also String uses char[] and some constructors also create copies of char[].

    – Puce
    7 hours ago











  • That's a good point actually, which leads me to believe @Turing85's answer is more correct without further clarification from the interviewers.

    – 17slim
    7 hours ago







3




3





Good answer, but for Java 8, I think you're overlooking the char[] object used by the implementation of StringBuilder.

– Andy Thomas
8 hours ago






Good answer, but for Java 8, I think you're overlooking the char[] object used by the implementation of StringBuilder.

– Andy Thomas
8 hours ago





1




1





stackoverflow.com/questions/12806739/…

– 17slim
7 hours ago





stackoverflow.com/questions/12806739/…

– 17slim
7 hours ago




4




4





@3limin4t0r "char[] is a primitive value" - Nope.

– Turing85
7 hours ago






@3limin4t0r "char[] is a primitive value" - Nope.

– Turing85
7 hours ago





2




2





@AndyThomas if you count the char[] you will end up with even more as also String uses char[] and some constructors also create copies of char[].

– Puce
7 hours ago





@AndyThomas if you count the char[] you will end up with even more as also String uses char[] and some constructors also create copies of char[].

– Puce
7 hours ago













That's a good point actually, which leads me to believe @Turing85's answer is more correct without further clarification from the interviewers.

– 17slim
7 hours ago





That's a good point actually, which leads me to believe @Turing85's answer is more correct without further clarification from the interviewers.

– 17slim
7 hours ago











2















It should be 5:



  • three for the three literals (assigned to str1, str2 and str3)


  • one for str1 + str2


  • one for (result from the previous operation) + str3 (assigned to str4)






share|improve this answer



























  • str1 + str2 will not be concatenated to a String in all Java since quite many years.

    – Puce
    8 hours ago















2















It should be 5:



  • three for the three literals (assigned to str1, str2 and str3)


  • one for str1 + str2


  • one for (result from the previous operation) + str3 (assigned to str4)






share|improve this answer



























  • str1 + str2 will not be concatenated to a String in all Java since quite many years.

    – Puce
    8 hours ago













2














2










2









It should be 5:



  • three for the three literals (assigned to str1, str2 and str3)


  • one for str1 + str2


  • one for (result from the previous operation) + str3 (assigned to str4)






share|improve this answer















It should be 5:



  • three for the three literals (assigned to str1, str2 and str3)


  • one for str1 + str2


  • one for (result from the previous operation) + str3 (assigned to str4)







share|improve this answer














share|improve this answer



share|improve this answer








edited 8 hours ago

























answered 8 hours ago









Laurenz AlbeLaurenz Albe

62.8k11 gold badges44 silver badges66 bronze badges




62.8k11 gold badges44 silver badges66 bronze badges















  • str1 + str2 will not be concatenated to a String in all Java since quite many years.

    – Puce
    8 hours ago

















  • str1 + str2 will not be concatenated to a String in all Java since quite many years.

    – Puce
    8 hours ago
















str1 + str2 will not be concatenated to a String in all Java since quite many years.

– Puce
8 hours ago





str1 + str2 will not be concatenated to a String in all Java since quite many years.

– Puce
8 hours ago











2















Any answer to your question will be dependant on the JVM implementation and the Java version currently being used. It's a stupid question to ask in an interview.



Java 1.8.0_201



On my machine, with Java 1.8.0_201, your snippet results in this bytecode



 L0
LINENUMBER 13 L0
LDC "First"
ASTORE 1
L1
LINENUMBER 14 L1
LDC "Second"
ASTORE 2
L2
LINENUMBER 15 L2
LDC "Third"
ASTORE 3
L3
LINENUMBER 16 L3
NEW java/lang/StringBuilder
DUP
INVOKESPECIAL java/lang/StringBuilder.<init> ()V
ALOAD 1
INVOKEVIRTUAL java/lang/StringBuilder.append (Ljava/lang/String;)Ljava/lang/StringBuilder;
ALOAD 2
INVOKEVIRTUAL java/lang/StringBuilder.append (Ljava/lang/String;)Ljava/lang/StringBuilder;
ALOAD 3
INVOKEVIRTUAL java/lang/StringBuilder.append (Ljava/lang/String;)Ljava/lang/StringBuilder;
INVOKEVIRTUAL java/lang/StringBuilder.toString ()Ljava/lang/String;
ASTORE 4
L4
LINENUMBER 19 L4
RETURN


which proves that 5 objects are being created (3 constant String literals, 1 StringBuilder, 1 dynamically produced String instance by StringBuilder#toString).



Java 12.0.2



On my machine, with Java 12.0.2, the bytecode is



 L0
LINENUMBER 13 L0
LDC "First"
ASTORE 1
L1
LINENUMBER 14 L1
LDC "Second"
ASTORE 2
L2
LINENUMBER 15 L2
LDC "Third"
ASTORE 3
L3
LINENUMBER 16 L3
ALOAD 1
ALOAD 2
ALOAD 3
INVOKEDYNAMIC makeConcatWithConstants(Ljava/lang/String;Ljava/lang/String;Ljava/lang/String;)Ljava/lang/String; [
// handle kind 0x6 : INVOKESTATIC
java/lang/invoke/StringConcatFactory.makeConcatWithConstants(Ljava/lang/invoke/MethodHandles$Lookup;Ljava/lang/String;Ljava/lang/invoke/MethodType;Ljava/lang/String;[Ljava/lang/Object;)Ljava/lang/invoke/CallSite;
// arguments:
"u0001u0001u0001"
]
ASTORE 4
L4
LINENUMBER 17 L4
RETURN


which magically changes "the correct answer" to 4 objects since there is no intermediate String[Buffer/Builder] involved.






share|improve this answer






















  • 1





    each string also contains a byte[], same for StringBuilder (and first one will also create a static byte[0]) and eventually (haven't checked) the StringBuilder can create a new, bigger buffer (if needed) (just proving your 2nd sentence true)

    – Carlos Heuberger
    7 hours ago












  • thinking about it, is LDC creating a (new) Object? or just using one? (the literal could have been used before)

    – Carlos Heuberger
    7 hours ago
















2















Any answer to your question will be dependant on the JVM implementation and the Java version currently being used. It's a stupid question to ask in an interview.



Java 1.8.0_201



On my machine, with Java 1.8.0_201, your snippet results in this bytecode



 L0
LINENUMBER 13 L0
LDC "First"
ASTORE 1
L1
LINENUMBER 14 L1
LDC "Second"
ASTORE 2
L2
LINENUMBER 15 L2
LDC "Third"
ASTORE 3
L3
LINENUMBER 16 L3
NEW java/lang/StringBuilder
DUP
INVOKESPECIAL java/lang/StringBuilder.<init> ()V
ALOAD 1
INVOKEVIRTUAL java/lang/StringBuilder.append (Ljava/lang/String;)Ljava/lang/StringBuilder;
ALOAD 2
INVOKEVIRTUAL java/lang/StringBuilder.append (Ljava/lang/String;)Ljava/lang/StringBuilder;
ALOAD 3
INVOKEVIRTUAL java/lang/StringBuilder.append (Ljava/lang/String;)Ljava/lang/StringBuilder;
INVOKEVIRTUAL java/lang/StringBuilder.toString ()Ljava/lang/String;
ASTORE 4
L4
LINENUMBER 19 L4
RETURN


which proves that 5 objects are being created (3 constant String literals, 1 StringBuilder, 1 dynamically produced String instance by StringBuilder#toString).



Java 12.0.2



On my machine, with Java 12.0.2, the bytecode is



 L0
LINENUMBER 13 L0
LDC "First"
ASTORE 1
L1
LINENUMBER 14 L1
LDC "Second"
ASTORE 2
L2
LINENUMBER 15 L2
LDC "Third"
ASTORE 3
L3
LINENUMBER 16 L3
ALOAD 1
ALOAD 2
ALOAD 3
INVOKEDYNAMIC makeConcatWithConstants(Ljava/lang/String;Ljava/lang/String;Ljava/lang/String;)Ljava/lang/String; [
// handle kind 0x6 : INVOKESTATIC
java/lang/invoke/StringConcatFactory.makeConcatWithConstants(Ljava/lang/invoke/MethodHandles$Lookup;Ljava/lang/String;Ljava/lang/invoke/MethodType;Ljava/lang/String;[Ljava/lang/Object;)Ljava/lang/invoke/CallSite;
// arguments:
"u0001u0001u0001"
]
ASTORE 4
L4
LINENUMBER 17 L4
RETURN


which magically changes "the correct answer" to 4 objects since there is no intermediate String[Buffer/Builder] involved.






share|improve this answer






















  • 1





    each string also contains a byte[], same for StringBuilder (and first one will also create a static byte[0]) and eventually (haven't checked) the StringBuilder can create a new, bigger buffer (if needed) (just proving your 2nd sentence true)

    – Carlos Heuberger
    7 hours ago












  • thinking about it, is LDC creating a (new) Object? or just using one? (the literal could have been used before)

    – Carlos Heuberger
    7 hours ago














2














2










2









Any answer to your question will be dependant on the JVM implementation and the Java version currently being used. It's a stupid question to ask in an interview.



Java 1.8.0_201



On my machine, with Java 1.8.0_201, your snippet results in this bytecode



 L0
LINENUMBER 13 L0
LDC "First"
ASTORE 1
L1
LINENUMBER 14 L1
LDC "Second"
ASTORE 2
L2
LINENUMBER 15 L2
LDC "Third"
ASTORE 3
L3
LINENUMBER 16 L3
NEW java/lang/StringBuilder
DUP
INVOKESPECIAL java/lang/StringBuilder.<init> ()V
ALOAD 1
INVOKEVIRTUAL java/lang/StringBuilder.append (Ljava/lang/String;)Ljava/lang/StringBuilder;
ALOAD 2
INVOKEVIRTUAL java/lang/StringBuilder.append (Ljava/lang/String;)Ljava/lang/StringBuilder;
ALOAD 3
INVOKEVIRTUAL java/lang/StringBuilder.append (Ljava/lang/String;)Ljava/lang/StringBuilder;
INVOKEVIRTUAL java/lang/StringBuilder.toString ()Ljava/lang/String;
ASTORE 4
L4
LINENUMBER 19 L4
RETURN


which proves that 5 objects are being created (3 constant String literals, 1 StringBuilder, 1 dynamically produced String instance by StringBuilder#toString).



Java 12.0.2



On my machine, with Java 12.0.2, the bytecode is



 L0
LINENUMBER 13 L0
LDC "First"
ASTORE 1
L1
LINENUMBER 14 L1
LDC "Second"
ASTORE 2
L2
LINENUMBER 15 L2
LDC "Third"
ASTORE 3
L3
LINENUMBER 16 L3
ALOAD 1
ALOAD 2
ALOAD 3
INVOKEDYNAMIC makeConcatWithConstants(Ljava/lang/String;Ljava/lang/String;Ljava/lang/String;)Ljava/lang/String; [
// handle kind 0x6 : INVOKESTATIC
java/lang/invoke/StringConcatFactory.makeConcatWithConstants(Ljava/lang/invoke/MethodHandles$Lookup;Ljava/lang/String;Ljava/lang/invoke/MethodType;Ljava/lang/String;[Ljava/lang/Object;)Ljava/lang/invoke/CallSite;
// arguments:
"u0001u0001u0001"
]
ASTORE 4
L4
LINENUMBER 17 L4
RETURN


which magically changes "the correct answer" to 4 objects since there is no intermediate String[Buffer/Builder] involved.






share|improve this answer















Any answer to your question will be dependant on the JVM implementation and the Java version currently being used. It's a stupid question to ask in an interview.



Java 1.8.0_201



On my machine, with Java 1.8.0_201, your snippet results in this bytecode



 L0
LINENUMBER 13 L0
LDC "First"
ASTORE 1
L1
LINENUMBER 14 L1
LDC "Second"
ASTORE 2
L2
LINENUMBER 15 L2
LDC "Third"
ASTORE 3
L3
LINENUMBER 16 L3
NEW java/lang/StringBuilder
DUP
INVOKESPECIAL java/lang/StringBuilder.<init> ()V
ALOAD 1
INVOKEVIRTUAL java/lang/StringBuilder.append (Ljava/lang/String;)Ljava/lang/StringBuilder;
ALOAD 2
INVOKEVIRTUAL java/lang/StringBuilder.append (Ljava/lang/String;)Ljava/lang/StringBuilder;
ALOAD 3
INVOKEVIRTUAL java/lang/StringBuilder.append (Ljava/lang/String;)Ljava/lang/StringBuilder;
INVOKEVIRTUAL java/lang/StringBuilder.toString ()Ljava/lang/String;
ASTORE 4
L4
LINENUMBER 19 L4
RETURN


which proves that 5 objects are being created (3 constant String literals, 1 StringBuilder, 1 dynamically produced String instance by StringBuilder#toString).



Java 12.0.2



On my machine, with Java 12.0.2, the bytecode is



 L0
LINENUMBER 13 L0
LDC "First"
ASTORE 1
L1
LINENUMBER 14 L1
LDC "Second"
ASTORE 2
L2
LINENUMBER 15 L2
LDC "Third"
ASTORE 3
L3
LINENUMBER 16 L3
ALOAD 1
ALOAD 2
ALOAD 3
INVOKEDYNAMIC makeConcatWithConstants(Ljava/lang/String;Ljava/lang/String;Ljava/lang/String;)Ljava/lang/String; [
// handle kind 0x6 : INVOKESTATIC
java/lang/invoke/StringConcatFactory.makeConcatWithConstants(Ljava/lang/invoke/MethodHandles$Lookup;Ljava/lang/String;Ljava/lang/invoke/MethodType;Ljava/lang/String;[Ljava/lang/Object;)Ljava/lang/invoke/CallSite;
// arguments:
"u0001u0001u0001"
]
ASTORE 4
L4
LINENUMBER 17 L4
RETURN


which magically changes "the correct answer" to 4 objects since there is no intermediate String[Buffer/Builder] involved.







share|improve this answer














share|improve this answer



share|improve this answer








edited 7 hours ago

























answered 7 hours ago









Andrew TobilkoAndrew Tobilko

32.9k10 gold badges53 silver badges102 bronze badges




32.9k10 gold badges53 silver badges102 bronze badges










  • 1





    each string also contains a byte[], same for StringBuilder (and first one will also create a static byte[0]) and eventually (haven't checked) the StringBuilder can create a new, bigger buffer (if needed) (just proving your 2nd sentence true)

    – Carlos Heuberger
    7 hours ago












  • thinking about it, is LDC creating a (new) Object? or just using one? (the literal could have been used before)

    – Carlos Heuberger
    7 hours ago













  • 1





    each string also contains a byte[], same for StringBuilder (and first one will also create a static byte[0]) and eventually (haven't checked) the StringBuilder can create a new, bigger buffer (if needed) (just proving your 2nd sentence true)

    – Carlos Heuberger
    7 hours ago












  • thinking about it, is LDC creating a (new) Object? or just using one? (the literal could have been used before)

    – Carlos Heuberger
    7 hours ago








1




1





each string also contains a byte[], same for StringBuilder (and first one will also create a static byte[0]) and eventually (haven't checked) the StringBuilder can create a new, bigger buffer (if needed) (just proving your 2nd sentence true)

– Carlos Heuberger
7 hours ago






each string also contains a byte[], same for StringBuilder (and first one will also create a static byte[0]) and eventually (haven't checked) the StringBuilder can create a new, bigger buffer (if needed) (just proving your 2nd sentence true)

– Carlos Heuberger
7 hours ago














thinking about it, is LDC creating a (new) Object? or just using one? (the literal could have been used before)

– Carlos Heuberger
7 hours ago






thinking about it, is LDC creating a (new) Object? or just using one? (the literal could have been used before)

– Carlos Heuberger
7 hours ago












0















Concatenation operation doesn't create those many String objects. It creates aStringBuilder and then appends the strings. So there may be 5 objects,
3 (variables) + 1 (sb) + 1 (Concatenated string).






share|improve this answer



























  • It might be 5.

    – Nexevis
    7 hours ago











  • As was stated many times, the JLS says that "a Java compiler may use the StringBuffer class or a similar technique to reduce the number of intermediate String objects that are created by evaluation of an expression." Emphasis is on may.

    – Turing85
    7 hours ago












  • @Turing85 agreed. Updated.

    – javaaddict
    7 hours ago















0















Concatenation operation doesn't create those many String objects. It creates aStringBuilder and then appends the strings. So there may be 5 objects,
3 (variables) + 1 (sb) + 1 (Concatenated string).






share|improve this answer



























  • It might be 5.

    – Nexevis
    7 hours ago











  • As was stated many times, the JLS says that "a Java compiler may use the StringBuffer class or a similar technique to reduce the number of intermediate String objects that are created by evaluation of an expression." Emphasis is on may.

    – Turing85
    7 hours ago












  • @Turing85 agreed. Updated.

    – javaaddict
    7 hours ago













0














0










0









Concatenation operation doesn't create those many String objects. It creates aStringBuilder and then appends the strings. So there may be 5 objects,
3 (variables) + 1 (sb) + 1 (Concatenated string).






share|improve this answer















Concatenation operation doesn't create those many String objects. It creates aStringBuilder and then appends the strings. So there may be 5 objects,
3 (variables) + 1 (sb) + 1 (Concatenated string).







share|improve this answer














share|improve this answer



share|improve this answer








edited 7 hours ago

























answered 7 hours ago









javaaddictjavaaddict

1371 silver badge8 bronze badges




1371 silver badge8 bronze badges















  • It might be 5.

    – Nexevis
    7 hours ago











  • As was stated many times, the JLS says that "a Java compiler may use the StringBuffer class or a similar technique to reduce the number of intermediate String objects that are created by evaluation of an expression." Emphasis is on may.

    – Turing85
    7 hours ago












  • @Turing85 agreed. Updated.

    – javaaddict
    7 hours ago

















  • It might be 5.

    – Nexevis
    7 hours ago











  • As was stated many times, the JLS says that "a Java compiler may use the StringBuffer class or a similar technique to reduce the number of intermediate String objects that are created by evaluation of an expression." Emphasis is on may.

    – Turing85
    7 hours ago












  • @Turing85 agreed. Updated.

    – javaaddict
    7 hours ago
















It might be 5.

– Nexevis
7 hours ago





It might be 5.

– Nexevis
7 hours ago













As was stated many times, the JLS says that "a Java compiler may use the StringBuffer class or a similar technique to reduce the number of intermediate String objects that are created by evaluation of an expression." Emphasis is on may.

– Turing85
7 hours ago






As was stated many times, the JLS says that "a Java compiler may use the StringBuffer class or a similar technique to reduce the number of intermediate String objects that are created by evaluation of an expression." Emphasis is on may.

– Turing85
7 hours ago














@Turing85 agreed. Updated.

– javaaddict
7 hours ago





@Turing85 agreed. Updated.

– javaaddict
7 hours ago










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