Replace this lambda with a method reference











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I have the following code. Sonar is complaining replace this lambda with a method reference.



Stream.iterate(0, i -> i + 1).limit(100).map(i -> Integer.toString(i));


If I replace it with it code below, it does not compile with compilation error: Type mismatch: cannot convert from Stream<Object> to <unknown>.



Stream.iterate(0, i -> i + 1).limit(100).map(Integer::toString);


How is Integer::toString converting Stream<Object> to <unknown>?










share|improve this question




























    up vote
    7
    down vote

    favorite












    I have the following code. Sonar is complaining replace this lambda with a method reference.



    Stream.iterate(0, i -> i + 1).limit(100).map(i -> Integer.toString(i));


    If I replace it with it code below, it does not compile with compilation error: Type mismatch: cannot convert from Stream<Object> to <unknown>.



    Stream.iterate(0, i -> i + 1).limit(100).map(Integer::toString);


    How is Integer::toString converting Stream<Object> to <unknown>?










    share|improve this question


























      up vote
      7
      down vote

      favorite









      up vote
      7
      down vote

      favorite











      I have the following code. Sonar is complaining replace this lambda with a method reference.



      Stream.iterate(0, i -> i + 1).limit(100).map(i -> Integer.toString(i));


      If I replace it with it code below, it does not compile with compilation error: Type mismatch: cannot convert from Stream<Object> to <unknown>.



      Stream.iterate(0, i -> i + 1).limit(100).map(Integer::toString);


      How is Integer::toString converting Stream<Object> to <unknown>?










      share|improve this question















      I have the following code. Sonar is complaining replace this lambda with a method reference.



      Stream.iterate(0, i -> i + 1).limit(100).map(i -> Integer.toString(i));


      If I replace it with it code below, it does not compile with compilation error: Type mismatch: cannot convert from Stream<Object> to <unknown>.



      Stream.iterate(0, i -> i + 1).limit(100).map(Integer::toString);


      How is Integer::toString converting Stream<Object> to <unknown>?







      java java-8 sonarqube java-stream






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 6 hours ago

























      asked 8 hours ago









      fastcodejava

      23.6k19109160




      23.6k19109160
























          4 Answers
          4






          active

          oldest

          votes

















          up vote
          8
          down vote













          It's ambiguous because the static and non-static toString() methods are both compatible with the functional signature Integer -> String. You can use String::valueOf instead.






          share|improve this answer




























            up vote
            5
            down vote













            You can't put Integer::toString because Integer has two implementations that fit to functional interface Function<Integer, String>, but you can use String::valueOf instead:



            Stream.iterate(0, i -> i + 1)
            .limit(100)
            .map(String::valueOf)
            .collect(Collectors.toList())





            share|improve this answer



















            • 1




              You can't put Integer::toString because it accepts int and your case you have Integer use. That's not correct. Lambdas can implicitly box and unbox.
              – shmosel
              7 hours ago












            • @shmosel then why does it IntStream.range(1, 100).mapToObj(Integer::toString).collect(Collectors.toList()) work?
              – vlad324
              7 hours ago












            • Because the int overload is more appropriate for a primitive stream.
              – shmosel
              7 hours ago












            • @shmosel yes, you are right about Lambdas can implicitly box and unbox
              – vlad324
              7 hours ago


















            up vote
            2
            down vote













            As mentioned by @shmosel already replacing the lambda with a method reference will lead to ambiguity as there's two toString method of the signarure:




            • String toString()

            • static String toString(int i)


            because the call to Stream.iterate(0, i -> i + 1) returns a Stream<Integer> when you call map with the method reference Integer::toString the compiler is not sure whether you meant to do Integer.toString(i) or i.toString() hence the compilation error.



            So here are other options to what's already been provided:



            instead of Stream.iterate you can use IntStream.iterate then call mapToObj:



            IntStream.iterate(0, i -> i + 1) // IntStream
            .limit(100) // IntStream
            .mapToObj(Integer::toString); // i1 -> Integer.toString(i1)


            Another thing suggested by intelliJ is that you can actually do:



            Stream.iterate(0, i -> i + 1) // Stream<Integer>
            .limit(100) // Stream<Integer>
            .map(Object::toString); // integer -> integer.toString()


            where Object::toString is equivalent to the lambda integer -> integer.toString()





            on another note, it's interesting that Sonar is suggesting to replace the lambda with a method reference in the code you've shown. intelliJ IDEA was smart enough not to suggest it.






            share|improve this answer






























              up vote
              0
              down vote













              I think that IntStream is better for your code:



              List<String> numbers = IntStream.range(0, 100)
              .mapToObj(String::valueOf)
              .collect(Collectors.toList());


              Or for your example do use String.valueOf to conver int -> String:



              List<String> numbers = Stream.iterate(0, i -> i + 1)
              .limit(100)
              .map(String::valueOf)
              .collect(Collectors.toList());





              share|improve this answer























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                4 Answers
                4






                active

                oldest

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                4 Answers
                4






                active

                oldest

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                active

                oldest

                votes






                active

                oldest

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                up vote
                8
                down vote













                It's ambiguous because the static and non-static toString() methods are both compatible with the functional signature Integer -> String. You can use String::valueOf instead.






                share|improve this answer

























                  up vote
                  8
                  down vote













                  It's ambiguous because the static and non-static toString() methods are both compatible with the functional signature Integer -> String. You can use String::valueOf instead.






                  share|improve this answer























                    up vote
                    8
                    down vote










                    up vote
                    8
                    down vote









                    It's ambiguous because the static and non-static toString() methods are both compatible with the functional signature Integer -> String. You can use String::valueOf instead.






                    share|improve this answer












                    It's ambiguous because the static and non-static toString() methods are both compatible with the functional signature Integer -> String. You can use String::valueOf instead.







                    share|improve this answer












                    share|improve this answer



                    share|improve this answer










                    answered 8 hours ago









                    shmosel

                    35.6k43892




                    35.6k43892
























                        up vote
                        5
                        down vote













                        You can't put Integer::toString because Integer has two implementations that fit to functional interface Function<Integer, String>, but you can use String::valueOf instead:



                        Stream.iterate(0, i -> i + 1)
                        .limit(100)
                        .map(String::valueOf)
                        .collect(Collectors.toList())





                        share|improve this answer



















                        • 1




                          You can't put Integer::toString because it accepts int and your case you have Integer use. That's not correct. Lambdas can implicitly box and unbox.
                          – shmosel
                          7 hours ago












                        • @shmosel then why does it IntStream.range(1, 100).mapToObj(Integer::toString).collect(Collectors.toList()) work?
                          – vlad324
                          7 hours ago












                        • Because the int overload is more appropriate for a primitive stream.
                          – shmosel
                          7 hours ago












                        • @shmosel yes, you are right about Lambdas can implicitly box and unbox
                          – vlad324
                          7 hours ago















                        up vote
                        5
                        down vote













                        You can't put Integer::toString because Integer has two implementations that fit to functional interface Function<Integer, String>, but you can use String::valueOf instead:



                        Stream.iterate(0, i -> i + 1)
                        .limit(100)
                        .map(String::valueOf)
                        .collect(Collectors.toList())





                        share|improve this answer



















                        • 1




                          You can't put Integer::toString because it accepts int and your case you have Integer use. That's not correct. Lambdas can implicitly box and unbox.
                          – shmosel
                          7 hours ago












                        • @shmosel then why does it IntStream.range(1, 100).mapToObj(Integer::toString).collect(Collectors.toList()) work?
                          – vlad324
                          7 hours ago












                        • Because the int overload is more appropriate for a primitive stream.
                          – shmosel
                          7 hours ago












                        • @shmosel yes, you are right about Lambdas can implicitly box and unbox
                          – vlad324
                          7 hours ago













                        up vote
                        5
                        down vote










                        up vote
                        5
                        down vote









                        You can't put Integer::toString because Integer has two implementations that fit to functional interface Function<Integer, String>, but you can use String::valueOf instead:



                        Stream.iterate(0, i -> i + 1)
                        .limit(100)
                        .map(String::valueOf)
                        .collect(Collectors.toList())





                        share|improve this answer














                        You can't put Integer::toString because Integer has two implementations that fit to functional interface Function<Integer, String>, but you can use String::valueOf instead:



                        Stream.iterate(0, i -> i + 1)
                        .limit(100)
                        .map(String::valueOf)
                        .collect(Collectors.toList())






                        share|improve this answer














                        share|improve this answer



                        share|improve this answer








                        edited 7 hours ago

























                        answered 8 hours ago









                        vlad324

                        53228




                        53228








                        • 1




                          You can't put Integer::toString because it accepts int and your case you have Integer use. That's not correct. Lambdas can implicitly box and unbox.
                          – shmosel
                          7 hours ago












                        • @shmosel then why does it IntStream.range(1, 100).mapToObj(Integer::toString).collect(Collectors.toList()) work?
                          – vlad324
                          7 hours ago












                        • Because the int overload is more appropriate for a primitive stream.
                          – shmosel
                          7 hours ago












                        • @shmosel yes, you are right about Lambdas can implicitly box and unbox
                          – vlad324
                          7 hours ago














                        • 1




                          You can't put Integer::toString because it accepts int and your case you have Integer use. That's not correct. Lambdas can implicitly box and unbox.
                          – shmosel
                          7 hours ago












                        • @shmosel then why does it IntStream.range(1, 100).mapToObj(Integer::toString).collect(Collectors.toList()) work?
                          – vlad324
                          7 hours ago












                        • Because the int overload is more appropriate for a primitive stream.
                          – shmosel
                          7 hours ago












                        • @shmosel yes, you are right about Lambdas can implicitly box and unbox
                          – vlad324
                          7 hours ago








                        1




                        1




                        You can't put Integer::toString because it accepts int and your case you have Integer use. That's not correct. Lambdas can implicitly box and unbox.
                        – shmosel
                        7 hours ago






                        You can't put Integer::toString because it accepts int and your case you have Integer use. That's not correct. Lambdas can implicitly box and unbox.
                        – shmosel
                        7 hours ago














                        @shmosel then why does it IntStream.range(1, 100).mapToObj(Integer::toString).collect(Collectors.toList()) work?
                        – vlad324
                        7 hours ago






                        @shmosel then why does it IntStream.range(1, 100).mapToObj(Integer::toString).collect(Collectors.toList()) work?
                        – vlad324
                        7 hours ago














                        Because the int overload is more appropriate for a primitive stream.
                        – shmosel
                        7 hours ago






                        Because the int overload is more appropriate for a primitive stream.
                        – shmosel
                        7 hours ago














                        @shmosel yes, you are right about Lambdas can implicitly box and unbox
                        – vlad324
                        7 hours ago




                        @shmosel yes, you are right about Lambdas can implicitly box and unbox
                        – vlad324
                        7 hours ago










                        up vote
                        2
                        down vote













                        As mentioned by @shmosel already replacing the lambda with a method reference will lead to ambiguity as there's two toString method of the signarure:




                        • String toString()

                        • static String toString(int i)


                        because the call to Stream.iterate(0, i -> i + 1) returns a Stream<Integer> when you call map with the method reference Integer::toString the compiler is not sure whether you meant to do Integer.toString(i) or i.toString() hence the compilation error.



                        So here are other options to what's already been provided:



                        instead of Stream.iterate you can use IntStream.iterate then call mapToObj:



                        IntStream.iterate(0, i -> i + 1) // IntStream
                        .limit(100) // IntStream
                        .mapToObj(Integer::toString); // i1 -> Integer.toString(i1)


                        Another thing suggested by intelliJ is that you can actually do:



                        Stream.iterate(0, i -> i + 1) // Stream<Integer>
                        .limit(100) // Stream<Integer>
                        .map(Object::toString); // integer -> integer.toString()


                        where Object::toString is equivalent to the lambda integer -> integer.toString()





                        on another note, it's interesting that Sonar is suggesting to replace the lambda with a method reference in the code you've shown. intelliJ IDEA was smart enough not to suggest it.






                        share|improve this answer



























                          up vote
                          2
                          down vote













                          As mentioned by @shmosel already replacing the lambda with a method reference will lead to ambiguity as there's two toString method of the signarure:




                          • String toString()

                          • static String toString(int i)


                          because the call to Stream.iterate(0, i -> i + 1) returns a Stream<Integer> when you call map with the method reference Integer::toString the compiler is not sure whether you meant to do Integer.toString(i) or i.toString() hence the compilation error.



                          So here are other options to what's already been provided:



                          instead of Stream.iterate you can use IntStream.iterate then call mapToObj:



                          IntStream.iterate(0, i -> i + 1) // IntStream
                          .limit(100) // IntStream
                          .mapToObj(Integer::toString); // i1 -> Integer.toString(i1)


                          Another thing suggested by intelliJ is that you can actually do:



                          Stream.iterate(0, i -> i + 1) // Stream<Integer>
                          .limit(100) // Stream<Integer>
                          .map(Object::toString); // integer -> integer.toString()


                          where Object::toString is equivalent to the lambda integer -> integer.toString()





                          on another note, it's interesting that Sonar is suggesting to replace the lambda with a method reference in the code you've shown. intelliJ IDEA was smart enough not to suggest it.






                          share|improve this answer

























                            up vote
                            2
                            down vote










                            up vote
                            2
                            down vote









                            As mentioned by @shmosel already replacing the lambda with a method reference will lead to ambiguity as there's two toString method of the signarure:




                            • String toString()

                            • static String toString(int i)


                            because the call to Stream.iterate(0, i -> i + 1) returns a Stream<Integer> when you call map with the method reference Integer::toString the compiler is not sure whether you meant to do Integer.toString(i) or i.toString() hence the compilation error.



                            So here are other options to what's already been provided:



                            instead of Stream.iterate you can use IntStream.iterate then call mapToObj:



                            IntStream.iterate(0, i -> i + 1) // IntStream
                            .limit(100) // IntStream
                            .mapToObj(Integer::toString); // i1 -> Integer.toString(i1)


                            Another thing suggested by intelliJ is that you can actually do:



                            Stream.iterate(0, i -> i + 1) // Stream<Integer>
                            .limit(100) // Stream<Integer>
                            .map(Object::toString); // integer -> integer.toString()


                            where Object::toString is equivalent to the lambda integer -> integer.toString()





                            on another note, it's interesting that Sonar is suggesting to replace the lambda with a method reference in the code you've shown. intelliJ IDEA was smart enough not to suggest it.






                            share|improve this answer














                            As mentioned by @shmosel already replacing the lambda with a method reference will lead to ambiguity as there's two toString method of the signarure:




                            • String toString()

                            • static String toString(int i)


                            because the call to Stream.iterate(0, i -> i + 1) returns a Stream<Integer> when you call map with the method reference Integer::toString the compiler is not sure whether you meant to do Integer.toString(i) or i.toString() hence the compilation error.



                            So here are other options to what's already been provided:



                            instead of Stream.iterate you can use IntStream.iterate then call mapToObj:



                            IntStream.iterate(0, i -> i + 1) // IntStream
                            .limit(100) // IntStream
                            .mapToObj(Integer::toString); // i1 -> Integer.toString(i1)


                            Another thing suggested by intelliJ is that you can actually do:



                            Stream.iterate(0, i -> i + 1) // Stream<Integer>
                            .limit(100) // Stream<Integer>
                            .map(Object::toString); // integer -> integer.toString()


                            where Object::toString is equivalent to the lambda integer -> integer.toString()





                            on another note, it's interesting that Sonar is suggesting to replace the lambda with a method reference in the code you've shown. intelliJ IDEA was smart enough not to suggest it.







                            share|improve this answer














                            share|improve this answer



                            share|improve this answer








                            edited 8 mins ago

























                            answered 1 hour ago









                            Aomine

                            35.3k62859




                            35.3k62859






















                                up vote
                                0
                                down vote













                                I think that IntStream is better for your code:



                                List<String> numbers = IntStream.range(0, 100)
                                .mapToObj(String::valueOf)
                                .collect(Collectors.toList());


                                Or for your example do use String.valueOf to conver int -> String:



                                List<String> numbers = Stream.iterate(0, i -> i + 1)
                                .limit(100)
                                .map(String::valueOf)
                                .collect(Collectors.toList());





                                share|improve this answer



























                                  up vote
                                  0
                                  down vote













                                  I think that IntStream is better for your code:



                                  List<String> numbers = IntStream.range(0, 100)
                                  .mapToObj(String::valueOf)
                                  .collect(Collectors.toList());


                                  Or for your example do use String.valueOf to conver int -> String:



                                  List<String> numbers = Stream.iterate(0, i -> i + 1)
                                  .limit(100)
                                  .map(String::valueOf)
                                  .collect(Collectors.toList());





                                  share|improve this answer

























                                    up vote
                                    0
                                    down vote










                                    up vote
                                    0
                                    down vote









                                    I think that IntStream is better for your code:



                                    List<String> numbers = IntStream.range(0, 100)
                                    .mapToObj(String::valueOf)
                                    .collect(Collectors.toList());


                                    Or for your example do use String.valueOf to conver int -> String:



                                    List<String> numbers = Stream.iterate(0, i -> i + 1)
                                    .limit(100)
                                    .map(String::valueOf)
                                    .collect(Collectors.toList());





                                    share|improve this answer














                                    I think that IntStream is better for your code:



                                    List<String> numbers = IntStream.range(0, 100)
                                    .mapToObj(String::valueOf)
                                    .collect(Collectors.toList());


                                    Or for your example do use String.valueOf to conver int -> String:



                                    List<String> numbers = Stream.iterate(0, i -> i + 1)
                                    .limit(100)
                                    .map(String::valueOf)
                                    .collect(Collectors.toList());






                                    share|improve this answer














                                    share|improve this answer



                                    share|improve this answer








                                    edited 50 mins ago

























                                    answered 56 mins ago









                                    oleg.cherednik

                                    4,9092916




                                    4,9092916






























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