Printing the letters from A to Z using Java stream












6














I have this code but it gives me an error: Type mismatch: cannot convert from int to Character.



Stream.iterate('a', i -> i + 1).limit(26).forEach(System.out::println);


Although it is fine to write int i = 'a';



I know I can write it like this but that seems like too much code for a simple task.



Stream.iterate('a', i -> (char)(i + 1)).limit(26).forEach(System.out::println);


Why is the Java type inference failing?






java java-8 char java-stream







share|improve this question




















  • 2




    Related stackoverflow.com/a/32424763/1746118
    – nullpointer
    1 hour ago
















6














I have this code but it gives me an error: Type mismatch: cannot convert from int to Character.



Stream.iterate('a', i -> i + 1).limit(26).forEach(System.out::println);


Although it is fine to write int i = 'a';



I know I can write it like this but that seems like too much code for a simple task.



Stream.iterate('a', i -> (char)(i + 1)).limit(26).forEach(System.out::println);


Why is the Java type inference failing?






java java-8 char java-stream







share|improve this question




















  • 2




    Related stackoverflow.com/a/32424763/1746118
    – nullpointer
    1 hour ago














6












6








6


1





I have this code but it gives me an error: Type mismatch: cannot convert from int to Character.



Stream.iterate('a', i -> i + 1).limit(26).forEach(System.out::println);


Although it is fine to write int i = 'a';



I know I can write it like this but that seems like too much code for a simple task.



Stream.iterate('a', i -> (char)(i + 1)).limit(26).forEach(System.out::println);


Why is the Java type inference failing?






java java-8 char java-stream







share|improve this question















I have this code but it gives me an error: Type mismatch: cannot convert from int to Character.



Stream.iterate('a', i -> i + 1).limit(26).forEach(System.out::println);


Although it is fine to write int i = 'a';



I know I can write it like this but that seems like too much code for a simple task.



Stream.iterate('a', i -> (char)(i + 1)).limit(26).forEach(System.out::println);


Why is the Java type inference failing?






java java-8 char java-stream




java java-8 char java-stream






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 1 hour ago

























asked 5 hours ago









fastcodejava

23.9k19109161




23.9k19109161








  • 2




    Related stackoverflow.com/a/32424763/1746118
    – nullpointer
    1 hour ago














  • 2




    Related stackoverflow.com/a/32424763/1746118
    – nullpointer
    1 hour ago








2




2




Related stackoverflow.com/a/32424763/1746118
– nullpointer
1 hour ago




Related stackoverflow.com/a/32424763/1746118
– nullpointer
1 hour ago












2 Answers
2






active

oldest

votes


















10














The reason why i -> i + 1 does not compile is because you're attempting to implicitly convert an int to a Character which the compiler cannot do itself alone.



In other words, you can think of Stream.iterate('a', i -> i + 1) as:



Stream.iterate('a', (Character i) -> {

       int i1 = i + 1;  

       return i1; // not possible 

});


As you have noted, explicitly casting to char solves it:



Stream.iterate('a', i -> (char)(i + 1))...


Btw this is better done as:



IntStream.rangeClosed('a', 'z').forEach(c -> System.out.println((char)c));


This is better because:




  1. No boxing overhead thus more efficient

  2. if you were to stop at say letter h with the use of iterate you'd have to do more brain processing than just entering h as the upper bound with rangeClosed because you'd need to find the number to truncate the infinite stream upon.


  3. iterate generates an infinite stream which again has more overhead than a finite one.


etc...






share|improve this answer































    7














    How about just:



    Stream.iterate('a', i -> ++i).limit(26).forEach(System.out::println);



    i -> i + 1 does not work because i is a Character and i + 1 causes an implicit narrowing conversion (JLS 5.1.3), which is not allowed. You can explicitly cast it as was shown. However ++i works because (From JLS 15.15.1):




    Before the addition, binary numeric promotion (§5.6.2) is performed on the value 1 and the value of the variable. If necessary, the sum is narrowed by a narrowing primitive conversion (§5.1.3) and/or subjected to boxing conversion (§5.1.7) to the type of the variable before it is stored.




    The ++ operator takes care of the narrowing conversion without us having to explicitly cast it






    share|improve this answer



















    • 1




      Good answer, would have been even better if you explain why ++i works and i + i doesn't.
      – fastcodejava
      2 hours ago






    • 1




      and much better if you could also answer Why is the Java type inference is failing part specifically :)
      – nullpointer
      2 hours ago






    • 1




      @fastcodejava I have edited my answer to try to explain.
      – GBlodgett
      1 hour ago






    • 2




      1 👏🏻 for awesome explanation @GBlodgett
      – Deadpool
      1 hour ago











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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

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    active

    oldest

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    active

    oldest

    votes









    10














    The reason why i -> i + 1 does not compile is because you're attempting to implicitly convert an int to a Character which the compiler cannot do itself alone.



    In other words, you can think of Stream.iterate('a', i -> i + 1) as:



    Stream.iterate('a', (Character i) -> {
    
       int i1 = i + 1;  
    
       return i1; // not possible 
    
});


    As you have noted, explicitly casting to char solves it:



    Stream.iterate('a', i -> (char)(i + 1))...


    Btw this is better done as:



    IntStream.rangeClosed('a', 'z').forEach(c -> System.out.println((char)c));


    This is better because:




    1. No boxing overhead thus more efficient

    2. if you were to stop at say letter h with the use of iterate you'd have to do more brain processing than just entering h as the upper bound with rangeClosed because you'd need to find the number to truncate the infinite stream upon.


    3. iterate generates an infinite stream which again has more overhead than a finite one.


    etc...






    share|improve this answer




























      10














      The reason why i -> i + 1 does not compile is because you're attempting to implicitly convert an int to a Character which the compiler cannot do itself alone.



      In other words, you can think of Stream.iterate('a', i -> i + 1) as:



      Stream.iterate('a', (Character i) -> {
      
       int i1 = i + 1;  
      
       return i1; // not possible 
      
});


      As you have noted, explicitly casting to char solves it:



      Stream.iterate('a', i -> (char)(i + 1))...


      Btw this is better done as:



      IntStream.rangeClosed('a', 'z').forEach(c -> System.out.println((char)c));


      This is better because:




      1. No boxing overhead thus more efficient

      2. if you were to stop at say letter h with the use of iterate you'd have to do more brain processing than just entering h as the upper bound with rangeClosed because you'd need to find the number to truncate the infinite stream upon.


      3. iterate generates an infinite stream which again has more overhead than a finite one.


      etc...






      share|improve this answer


























        10












        10








        10






        The reason why i -> i + 1 does not compile is because you're attempting to implicitly convert an int to a Character which the compiler cannot do itself alone.



        In other words, you can think of Stream.iterate('a', i -> i + 1) as:



        Stream.iterate('a', (Character i) -> {
        
       int i1 = i + 1;  
        
       return i1; // not possible 
        
});


        As you have noted, explicitly casting to char solves it:



        Stream.iterate('a', i -> (char)(i + 1))...


        Btw this is better done as:



        IntStream.rangeClosed('a', 'z').forEach(c -> System.out.println((char)c));


        This is better because:




        1. No boxing overhead thus more efficient

        2. if you were to stop at say letter h with the use of iterate you'd have to do more brain processing than just entering h as the upper bound with rangeClosed because you'd need to find the number to truncate the infinite stream upon.


        3. iterate generates an infinite stream which again has more overhead than a finite one.


        etc...






        share|improve this answer














        The reason why i -> i + 1 does not compile is because you're attempting to implicitly convert an int to a Character which the compiler cannot do itself alone.



        In other words, you can think of Stream.iterate('a', i -> i + 1) as:



        Stream.iterate('a', (Character i) -> {
        
       int i1 = i + 1;  
        
       return i1; // not possible 
        
});


        As you have noted, explicitly casting to char solves it:



        Stream.iterate('a', i -> (char)(i + 1))...


        Btw this is better done as:



        IntStream.rangeClosed('a', 'z').forEach(c -> System.out.println((char)c));


        This is better because:




        1. No boxing overhead thus more efficient

        2. if you were to stop at say letter h with the use of iterate you'd have to do more brain processing than just entering h as the upper bound with rangeClosed because you'd need to find the number to truncate the infinite stream upon.


        3. iterate generates an infinite stream which again has more overhead than a finite one.


        etc...







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited 4 hours ago

























        answered 5 hours ago









        Aomine

        39.1k73367




        39.1k73367

























            7














            How about just:



            Stream.iterate('a', i -> ++i).limit(26).forEach(System.out::println);



            i -> i + 1 does not work because i is a Character and i + 1 causes an implicit narrowing conversion (JLS 5.1.3), which is not allowed. You can explicitly cast it as was shown. However ++i works because (From JLS 15.15.1):




            Before the addition, binary numeric promotion (§5.6.2) is performed on the value 1 and the value of the variable. If necessary, the sum is narrowed by a narrowing primitive conversion (§5.1.3) and/or subjected to boxing conversion (§5.1.7) to the type of the variable before it is stored.




            The ++ operator takes care of the narrowing conversion without us having to explicitly cast it






            share|improve this answer



















            • 1




              Good answer, would have been even better if you explain why ++i works and i + i doesn't.
              – fastcodejava
              2 hours ago






            • 1




              and much better if you could also answer Why is the Java type inference is failing part specifically :)
              – nullpointer
              2 hours ago






            • 1




              @fastcodejava I have edited my answer to try to explain.
              – GBlodgett
              1 hour ago






            • 2




              1 👏🏻 for awesome explanation @GBlodgett
              – Deadpool
              1 hour ago
















            7














            How about just:



            Stream.iterate('a', i -> ++i).limit(26).forEach(System.out::println);



            i -> i + 1 does not work because i is a Character and i + 1 causes an implicit narrowing conversion (JLS 5.1.3), which is not allowed. You can explicitly cast it as was shown. However ++i works because (From JLS 15.15.1):




            Before the addition, binary numeric promotion (§5.6.2) is performed on the value 1 and the value of the variable. If necessary, the sum is narrowed by a narrowing primitive conversion (§5.1.3) and/or subjected to boxing conversion (§5.1.7) to the type of the variable before it is stored.




            The ++ operator takes care of the narrowing conversion without us having to explicitly cast it






            share|improve this answer



















            • 1




              Good answer, would have been even better if you explain why ++i works and i + i doesn't.
              – fastcodejava
              2 hours ago






            • 1




              and much better if you could also answer Why is the Java type inference is failing part specifically :)
              – nullpointer
              2 hours ago






            • 1




              @fastcodejava I have edited my answer to try to explain.
              – GBlodgett
              1 hour ago






            • 2




              1 👏🏻 for awesome explanation @GBlodgett
              – Deadpool
              1 hour ago














            7












            7








            7






            How about just:



            Stream.iterate('a', i -> ++i).limit(26).forEach(System.out::println);



            i -> i + 1 does not work because i is a Character and i + 1 causes an implicit narrowing conversion (JLS 5.1.3), which is not allowed. You can explicitly cast it as was shown. However ++i works because (From JLS 15.15.1):




            Before the addition, binary numeric promotion (§5.6.2) is performed on the value 1 and the value of the variable. If necessary, the sum is narrowed by a narrowing primitive conversion (§5.1.3) and/or subjected to boxing conversion (§5.1.7) to the type of the variable before it is stored.




            The ++ operator takes care of the narrowing conversion without us having to explicitly cast it






            share|improve this answer














            How about just:



            Stream.iterate('a', i -> ++i).limit(26).forEach(System.out::println);



            i -> i + 1 does not work because i is a Character and i + 1 causes an implicit narrowing conversion (JLS 5.1.3), which is not allowed. You can explicitly cast it as was shown. However ++i works because (From JLS 15.15.1):




            Before the addition, binary numeric promotion (§5.6.2) is performed on the value 1 and the value of the variable. If necessary, the sum is narrowed by a narrowing primitive conversion (§5.1.3) and/or subjected to boxing conversion (§5.1.7) to the type of the variable before it is stored.




            The ++ operator takes care of the narrowing conversion without us having to explicitly cast it







            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited 1 hour ago

























            answered 5 hours ago









            GBlodgett

            9,10641632




            9,10641632








            • 1




              Good answer, would have been even better if you explain why ++i works and i + i doesn't.
              – fastcodejava
              2 hours ago






            • 1




              and much better if you could also answer Why is the Java type inference is failing part specifically :)
              – nullpointer
              2 hours ago






            • 1




              @fastcodejava I have edited my answer to try to explain.
              – GBlodgett
              1 hour ago






            • 2




              1 👏🏻 for awesome explanation @GBlodgett
              – Deadpool
              1 hour ago














            • 1




              Good answer, would have been even better if you explain why ++i works and i + i doesn't.
              – fastcodejava
              2 hours ago






            • 1




              and much better if you could also answer Why is the Java type inference is failing part specifically :)
              – nullpointer
              2 hours ago






            • 1




              @fastcodejava I have edited my answer to try to explain.
              – GBlodgett
              1 hour ago






            • 2




              1 👏🏻 for awesome explanation @GBlodgett
              – Deadpool
              1 hour ago








            1




            1




            Good answer, would have been even better if you explain why ++i works and i + i doesn't.
            – fastcodejava
            2 hours ago




            Good answer, would have been even better if you explain why ++i works and i + i doesn't.
            – fastcodejava
            2 hours ago




            1




            1




            and much better if you could also answer Why is the Java type inference is failing part specifically :)
            – nullpointer
            2 hours ago




            and much better if you could also answer Why is the Java type inference is failing part specifically :)
            – nullpointer
            2 hours ago




            1




            1




            @fastcodejava I have edited my answer to try to explain.
            – GBlodgett
            1 hour ago




            @fastcodejava I have edited my answer to try to explain.
            – GBlodgett
            1 hour ago




            2




            2




            1 👏🏻 for awesome explanation @GBlodgett
            – Deadpool
            1 hour ago




            1 👏🏻 for awesome explanation @GBlodgett
            – Deadpool
            1 hour ago


















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