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Why is a Java array index expression evaluated before checking if the array reference expression is null?


What are the rules for evaluation order in Java?Is the array index or the assigned value evaluated first?Checking for a null int value from a Java ResultSetIs null check needed before calling instanceof?Checking if a string is empty or null in JavaWhy can I throw null in Java?Printing an array with no elements?How can I use an array within a method that was called in as a parameter for a different constructor initializer?? (Java)Runtime evaluation of expressions in Java method referencesNullPointerException when switching activites and using a global arrayWhy does array[idx++]+=“a” increase idx once in Java 8 but twice in Java 9 and 10?What is the reason behind null checks in method reference expression evaluation?













33















According to the JLS, runtime evaluation of an array access expression behaves as follows:




  1. First, the array reference expression is evaluated. If this
    evaluation completes abruptly, then the array access completes
    abruptly for the same reason and the index expression is not
    evaluated.

  2. Otherwise, the index expression is evaluated. If this
    evaluation completes abruptly, then the array access completes
    abruptly for the same reason.

  3. Otherwise, if the value of the array
    reference expression is null, then a NullPointerException is thrown.


So this code will print: java.lang.NullPointerException, index=2



class Test3 {
public static void main(String[] args) {
int index = 1;
try {
nada()[index = 2]++;
} catch (Exception e) {
System.out.println(e + ", index=" + index);
}
}

static int[] nada() {
return null;
}
}


The question is: for what reason do we need to first evaluate the index = 2 expression and not just throw the NullPointerException once the array reference is evaluated to null? Or in other words - why is the order 1,2,3 and not 1,3,2?










share|improve this question




















  • 1





    Firstly you have to initialize an array and secondly the priority of equals sign is high that's why its given null pointer exception.

    – Ammar Ali
    Mar 14 at 11:10






  • 8





    Asking on SO why the JLS is written the way it is will not give you any good answer unless it maybe comes from one of the designers of the Java language.

    – Seelenvirtuose
    Mar 14 at 11:17








  • 4





    They had to choose something, and both options can produce "unexpected" scenarios. (I.e. scenarios that behave in a not very intuitive way). The option that was chosen seems to be more in line with how expression evaluation works in other parts of Java.

    – biziclop
    Mar 14 at 11:35






  • 2





    … which in turn is a duplicate of Is the array index or the assigned value evaluated first?

    – fantaghirocco
    Mar 14 at 13:22













  • @fantaghirocco i didn't ask about how java evaluates an array index expression, it's described in the question. I just wanted to know what the reason behind that's behavior.

    – Andrei Nepsha
    Mar 14 at 13:43


















33















According to the JLS, runtime evaluation of an array access expression behaves as follows:




  1. First, the array reference expression is evaluated. If this
    evaluation completes abruptly, then the array access completes
    abruptly for the same reason and the index expression is not
    evaluated.

  2. Otherwise, the index expression is evaluated. If this
    evaluation completes abruptly, then the array access completes
    abruptly for the same reason.

  3. Otherwise, if the value of the array
    reference expression is null, then a NullPointerException is thrown.


So this code will print: java.lang.NullPointerException, index=2



class Test3 {
public static void main(String[] args) {
int index = 1;
try {
nada()[index = 2]++;
} catch (Exception e) {
System.out.println(e + ", index=" + index);
}
}

static int[] nada() {
return null;
}
}


The question is: for what reason do we need to first evaluate the index = 2 expression and not just throw the NullPointerException once the array reference is evaluated to null? Or in other words - why is the order 1,2,3 and not 1,3,2?










share|improve this question




















  • 1





    Firstly you have to initialize an array and secondly the priority of equals sign is high that's why its given null pointer exception.

    – Ammar Ali
    Mar 14 at 11:10






  • 8





    Asking on SO why the JLS is written the way it is will not give you any good answer unless it maybe comes from one of the designers of the Java language.

    – Seelenvirtuose
    Mar 14 at 11:17








  • 4





    They had to choose something, and both options can produce "unexpected" scenarios. (I.e. scenarios that behave in a not very intuitive way). The option that was chosen seems to be more in line with how expression evaluation works in other parts of Java.

    – biziclop
    Mar 14 at 11:35






  • 2





    … which in turn is a duplicate of Is the array index or the assigned value evaluated first?

    – fantaghirocco
    Mar 14 at 13:22













  • @fantaghirocco i didn't ask about how java evaluates an array index expression, it's described in the question. I just wanted to know what the reason behind that's behavior.

    – Andrei Nepsha
    Mar 14 at 13:43
















33












33








33


4






According to the JLS, runtime evaluation of an array access expression behaves as follows:




  1. First, the array reference expression is evaluated. If this
    evaluation completes abruptly, then the array access completes
    abruptly for the same reason and the index expression is not
    evaluated.

  2. Otherwise, the index expression is evaluated. If this
    evaluation completes abruptly, then the array access completes
    abruptly for the same reason.

  3. Otherwise, if the value of the array
    reference expression is null, then a NullPointerException is thrown.


So this code will print: java.lang.NullPointerException, index=2



class Test3 {
public static void main(String[] args) {
int index = 1;
try {
nada()[index = 2]++;
} catch (Exception e) {
System.out.println(e + ", index=" + index);
}
}

static int[] nada() {
return null;
}
}


The question is: for what reason do we need to first evaluate the index = 2 expression and not just throw the NullPointerException once the array reference is evaluated to null? Or in other words - why is the order 1,2,3 and not 1,3,2?










share|improve this question
















According to the JLS, runtime evaluation of an array access expression behaves as follows:




  1. First, the array reference expression is evaluated. If this
    evaluation completes abruptly, then the array access completes
    abruptly for the same reason and the index expression is not
    evaluated.

  2. Otherwise, the index expression is evaluated. If this
    evaluation completes abruptly, then the array access completes
    abruptly for the same reason.

  3. Otherwise, if the value of the array
    reference expression is null, then a NullPointerException is thrown.


So this code will print: java.lang.NullPointerException, index=2



class Test3 {
public static void main(String[] args) {
int index = 1;
try {
nada()[index = 2]++;
} catch (Exception e) {
System.out.println(e + ", index=" + index);
}
}

static int[] nada() {
return null;
}
}


The question is: for what reason do we need to first evaluate the index = 2 expression and not just throw the NullPointerException once the array reference is evaluated to null? Or in other words - why is the order 1,2,3 and not 1,3,2?







java language-lawyer






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Mar 14 at 16:09









senseiwu

2,01411333




2,01411333










asked Mar 14 at 11:00









Andrei NepshaAndrei Nepsha

17119




17119








  • 1





    Firstly you have to initialize an array and secondly the priority of equals sign is high that's why its given null pointer exception.

    – Ammar Ali
    Mar 14 at 11:10






  • 8





    Asking on SO why the JLS is written the way it is will not give you any good answer unless it maybe comes from one of the designers of the Java language.

    – Seelenvirtuose
    Mar 14 at 11:17








  • 4





    They had to choose something, and both options can produce "unexpected" scenarios. (I.e. scenarios that behave in a not very intuitive way). The option that was chosen seems to be more in line with how expression evaluation works in other parts of Java.

    – biziclop
    Mar 14 at 11:35






  • 2





    … which in turn is a duplicate of Is the array index or the assigned value evaluated first?

    – fantaghirocco
    Mar 14 at 13:22













  • @fantaghirocco i didn't ask about how java evaluates an array index expression, it's described in the question. I just wanted to know what the reason behind that's behavior.

    – Andrei Nepsha
    Mar 14 at 13:43
















  • 1





    Firstly you have to initialize an array and secondly the priority of equals sign is high that's why its given null pointer exception.

    – Ammar Ali
    Mar 14 at 11:10






  • 8





    Asking on SO why the JLS is written the way it is will not give you any good answer unless it maybe comes from one of the designers of the Java language.

    – Seelenvirtuose
    Mar 14 at 11:17








  • 4





    They had to choose something, and both options can produce "unexpected" scenarios. (I.e. scenarios that behave in a not very intuitive way). The option that was chosen seems to be more in line with how expression evaluation works in other parts of Java.

    – biziclop
    Mar 14 at 11:35






  • 2





    … which in turn is a duplicate of Is the array index or the assigned value evaluated first?

    – fantaghirocco
    Mar 14 at 13:22













  • @fantaghirocco i didn't ask about how java evaluates an array index expression, it's described in the question. I just wanted to know what the reason behind that's behavior.

    – Andrei Nepsha
    Mar 14 at 13:43










1




1





Firstly you have to initialize an array and secondly the priority of equals sign is high that's why its given null pointer exception.

– Ammar Ali
Mar 14 at 11:10





Firstly you have to initialize an array and secondly the priority of equals sign is high that's why its given null pointer exception.

– Ammar Ali
Mar 14 at 11:10




8




8





Asking on SO why the JLS is written the way it is will not give you any good answer unless it maybe comes from one of the designers of the Java language.

– Seelenvirtuose
Mar 14 at 11:17







Asking on SO why the JLS is written the way it is will not give you any good answer unless it maybe comes from one of the designers of the Java language.

– Seelenvirtuose
Mar 14 at 11:17






4




4





They had to choose something, and both options can produce "unexpected" scenarios. (I.e. scenarios that behave in a not very intuitive way). The option that was chosen seems to be more in line with how expression evaluation works in other parts of Java.

– biziclop
Mar 14 at 11:35





They had to choose something, and both options can produce "unexpected" scenarios. (I.e. scenarios that behave in a not very intuitive way). The option that was chosen seems to be more in line with how expression evaluation works in other parts of Java.

– biziclop
Mar 14 at 11:35




2




2





… which in turn is a duplicate of Is the array index or the assigned value evaluated first?

– fantaghirocco
Mar 14 at 13:22







… which in turn is a duplicate of Is the array index or the assigned value evaluated first?

– fantaghirocco
Mar 14 at 13:22















@fantaghirocco i didn't ask about how java evaluates an array index expression, it's described in the question. I just wanted to know what the reason behind that's behavior.

– Andrei Nepsha
Mar 14 at 13:43







@fantaghirocco i didn't ask about how java evaluates an array index expression, it's described in the question. I just wanted to know what the reason behind that's behavior.

– Andrei Nepsha
Mar 14 at 13:43














4 Answers
4






active

oldest

votes


















36














An array access expression has two sub-expressions:




An array access expression contains two subexpressions, the array reference expression (before the left bracket) and the index expression (within the brackets).




The two sub-expressions are evaluated before the array access expression itself, in order to evaluate the expression.



After evaluating the two sub-expressions



nada()[index = 2]++;


becomes



null[2]++;


Only now the expression is evaluated and the NullPointerException is thrown.



This is consistent with the evaluation of most expressions in Java (the only counter examples I can think of are short circuiting operators such as && and ||).



For example, if you make the following method call:



firstMethod().secondMethod(i = 2);


First you evaluate firstMethod() and i = 2, and only later you throw NullPointerException if firstMethod() evaluated to null.






share|improve this answer































    13














    This is because in the generated bytecode there are no explicit null checks.



    nada()[index = 2]++;


    is translated into the following byte code:



    // evaluate the array reference expression
    INVOKESTATIC Test3.nada ()[I
    // evaluate the index expression
    ICONST_2
    DUP
    ISTORE 1
    // access the array
    // if the array reference expression was null, the IALOAD operation will throw a null pointer exception
    DUP2
    IALOAD
    ICONST_1
    IADD
    IASTORE





    share|improve this answer































      7














      The basic byte code operations are (for an int[])



      ALOAD array_address
      ILOAD index
      IALOAD array_element_retrieval


      The IALOAD does the null pointer check. In reality the code is a bit more elaborate:




      1. calculate array address

      2. calculate index

      3. IALOAD


      So the answer is: it would need an extra checking operation after the array address is loaded, in anticipation of the array access.



      Behavior by straight implementation.






      share|improve this answer































        7














        The decision may be partially be rooted in performance.



        In order to know that index = 2 is not going to be required, we would have to first evaluate nada() and then check whether it was null. We would then branch on the result of this condition, and decide whether or not to evaluate the array index expression.



        Every perfectly valid array index expression would be made slower by one additional operation, just for the sake of saving code - code that is going to throw an exception anyway - from evaluating one expression unnecessarily.



        It is an optimistic approach which works better in the majority of cases.






        share|improve this answer























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          4 Answers
          4






          active

          oldest

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          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          36














          An array access expression has two sub-expressions:




          An array access expression contains two subexpressions, the array reference expression (before the left bracket) and the index expression (within the brackets).




          The two sub-expressions are evaluated before the array access expression itself, in order to evaluate the expression.



          After evaluating the two sub-expressions



          nada()[index = 2]++;


          becomes



          null[2]++;


          Only now the expression is evaluated and the NullPointerException is thrown.



          This is consistent with the evaluation of most expressions in Java (the only counter examples I can think of are short circuiting operators such as && and ||).



          For example, if you make the following method call:



          firstMethod().secondMethod(i = 2);


          First you evaluate firstMethod() and i = 2, and only later you throw NullPointerException if firstMethod() evaluated to null.






          share|improve this answer




























            36














            An array access expression has two sub-expressions:




            An array access expression contains two subexpressions, the array reference expression (before the left bracket) and the index expression (within the brackets).




            The two sub-expressions are evaluated before the array access expression itself, in order to evaluate the expression.



            After evaluating the two sub-expressions



            nada()[index = 2]++;


            becomes



            null[2]++;


            Only now the expression is evaluated and the NullPointerException is thrown.



            This is consistent with the evaluation of most expressions in Java (the only counter examples I can think of are short circuiting operators such as && and ||).



            For example, if you make the following method call:



            firstMethod().secondMethod(i = 2);


            First you evaluate firstMethod() and i = 2, and only later you throw NullPointerException if firstMethod() evaluated to null.






            share|improve this answer


























              36












              36








              36







              An array access expression has two sub-expressions:




              An array access expression contains two subexpressions, the array reference expression (before the left bracket) and the index expression (within the brackets).




              The two sub-expressions are evaluated before the array access expression itself, in order to evaluate the expression.



              After evaluating the two sub-expressions



              nada()[index = 2]++;


              becomes



              null[2]++;


              Only now the expression is evaluated and the NullPointerException is thrown.



              This is consistent with the evaluation of most expressions in Java (the only counter examples I can think of are short circuiting operators such as && and ||).



              For example, if you make the following method call:



              firstMethod().secondMethod(i = 2);


              First you evaluate firstMethod() and i = 2, and only later you throw NullPointerException if firstMethod() evaluated to null.






              share|improve this answer













              An array access expression has two sub-expressions:




              An array access expression contains two subexpressions, the array reference expression (before the left bracket) and the index expression (within the brackets).




              The two sub-expressions are evaluated before the array access expression itself, in order to evaluate the expression.



              After evaluating the two sub-expressions



              nada()[index = 2]++;


              becomes



              null[2]++;


              Only now the expression is evaluated and the NullPointerException is thrown.



              This is consistent with the evaluation of most expressions in Java (the only counter examples I can think of are short circuiting operators such as && and ||).



              For example, if you make the following method call:



              firstMethod().secondMethod(i = 2);


              First you evaluate firstMethod() and i = 2, and only later you throw NullPointerException if firstMethod() evaluated to null.







              share|improve this answer












              share|improve this answer



              share|improve this answer










              answered Mar 14 at 11:30









              EranEran

              290k37475561




              290k37475561

























                  13














                  This is because in the generated bytecode there are no explicit null checks.



                  nada()[index = 2]++;


                  is translated into the following byte code:



                  // evaluate the array reference expression
                  INVOKESTATIC Test3.nada ()[I
                  // evaluate the index expression
                  ICONST_2
                  DUP
                  ISTORE 1
                  // access the array
                  // if the array reference expression was null, the IALOAD operation will throw a null pointer exception
                  DUP2
                  IALOAD
                  ICONST_1
                  IADD
                  IASTORE





                  share|improve this answer




























                    13














                    This is because in the generated bytecode there are no explicit null checks.



                    nada()[index = 2]++;


                    is translated into the following byte code:



                    // evaluate the array reference expression
                    INVOKESTATIC Test3.nada ()[I
                    // evaluate the index expression
                    ICONST_2
                    DUP
                    ISTORE 1
                    // access the array
                    // if the array reference expression was null, the IALOAD operation will throw a null pointer exception
                    DUP2
                    IALOAD
                    ICONST_1
                    IADD
                    IASTORE





                    share|improve this answer


























                      13












                      13








                      13







                      This is because in the generated bytecode there are no explicit null checks.



                      nada()[index = 2]++;


                      is translated into the following byte code:



                      // evaluate the array reference expression
                      INVOKESTATIC Test3.nada ()[I
                      // evaluate the index expression
                      ICONST_2
                      DUP
                      ISTORE 1
                      // access the array
                      // if the array reference expression was null, the IALOAD operation will throw a null pointer exception
                      DUP2
                      IALOAD
                      ICONST_1
                      IADD
                      IASTORE





                      share|improve this answer













                      This is because in the generated bytecode there are no explicit null checks.



                      nada()[index = 2]++;


                      is translated into the following byte code:



                      // evaluate the array reference expression
                      INVOKESTATIC Test3.nada ()[I
                      // evaluate the index expression
                      ICONST_2
                      DUP
                      ISTORE 1
                      // access the array
                      // if the array reference expression was null, the IALOAD operation will throw a null pointer exception
                      DUP2
                      IALOAD
                      ICONST_1
                      IADD
                      IASTORE






                      share|improve this answer












                      share|improve this answer



                      share|improve this answer










                      answered Mar 14 at 11:27









                      Thomas KlägerThomas Kläger

                      7,0182819




                      7,0182819























                          7














                          The basic byte code operations are (for an int[])



                          ALOAD array_address
                          ILOAD index
                          IALOAD array_element_retrieval


                          The IALOAD does the null pointer check. In reality the code is a bit more elaborate:




                          1. calculate array address

                          2. calculate index

                          3. IALOAD


                          So the answer is: it would need an extra checking operation after the array address is loaded, in anticipation of the array access.



                          Behavior by straight implementation.






                          share|improve this answer




























                            7














                            The basic byte code operations are (for an int[])



                            ALOAD array_address
                            ILOAD index
                            IALOAD array_element_retrieval


                            The IALOAD does the null pointer check. In reality the code is a bit more elaborate:




                            1. calculate array address

                            2. calculate index

                            3. IALOAD


                            So the answer is: it would need an extra checking operation after the array address is loaded, in anticipation of the array access.



                            Behavior by straight implementation.






                            share|improve this answer


























                              7












                              7








                              7







                              The basic byte code operations are (for an int[])



                              ALOAD array_address
                              ILOAD index
                              IALOAD array_element_retrieval


                              The IALOAD does the null pointer check. In reality the code is a bit more elaborate:




                              1. calculate array address

                              2. calculate index

                              3. IALOAD


                              So the answer is: it would need an extra checking operation after the array address is loaded, in anticipation of the array access.



                              Behavior by straight implementation.






                              share|improve this answer













                              The basic byte code operations are (for an int[])



                              ALOAD array_address
                              ILOAD index
                              IALOAD array_element_retrieval


                              The IALOAD does the null pointer check. In reality the code is a bit more elaborate:




                              1. calculate array address

                              2. calculate index

                              3. IALOAD


                              So the answer is: it would need an extra checking operation after the array address is loaded, in anticipation of the array access.



                              Behavior by straight implementation.







                              share|improve this answer












                              share|improve this answer



                              share|improve this answer










                              answered Mar 14 at 11:35









                              Joop EggenJoop Eggen

                              78.4k755105




                              78.4k755105























                                  7














                                  The decision may be partially be rooted in performance.



                                  In order to know that index = 2 is not going to be required, we would have to first evaluate nada() and then check whether it was null. We would then branch on the result of this condition, and decide whether or not to evaluate the array index expression.



                                  Every perfectly valid array index expression would be made slower by one additional operation, just for the sake of saving code - code that is going to throw an exception anyway - from evaluating one expression unnecessarily.



                                  It is an optimistic approach which works better in the majority of cases.






                                  share|improve this answer




























                                    7














                                    The decision may be partially be rooted in performance.



                                    In order to know that index = 2 is not going to be required, we would have to first evaluate nada() and then check whether it was null. We would then branch on the result of this condition, and decide whether or not to evaluate the array index expression.



                                    Every perfectly valid array index expression would be made slower by one additional operation, just for the sake of saving code - code that is going to throw an exception anyway - from evaluating one expression unnecessarily.



                                    It is an optimistic approach which works better in the majority of cases.






                                    share|improve this answer


























                                      7












                                      7








                                      7







                                      The decision may be partially be rooted in performance.



                                      In order to know that index = 2 is not going to be required, we would have to first evaluate nada() and then check whether it was null. We would then branch on the result of this condition, and decide whether or not to evaluate the array index expression.



                                      Every perfectly valid array index expression would be made slower by one additional operation, just for the sake of saving code - code that is going to throw an exception anyway - from evaluating one expression unnecessarily.



                                      It is an optimistic approach which works better in the majority of cases.






                                      share|improve this answer













                                      The decision may be partially be rooted in performance.



                                      In order to know that index = 2 is not going to be required, we would have to first evaluate nada() and then check whether it was null. We would then branch on the result of this condition, and decide whether or not to evaluate the array index expression.



                                      Every perfectly valid array index expression would be made slower by one additional operation, just for the sake of saving code - code that is going to throw an exception anyway - from evaluating one expression unnecessarily.



                                      It is an optimistic approach which works better in the majority of cases.







                                      share|improve this answer












                                      share|improve this answer



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                                      answered Mar 14 at 11:48









                                      MichaelMichael

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