Printing the letters from A to Z using a Java stream












10














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




















  • 4




    Related stackoverflow.com/a/32424763/1746118
    – nullpointer
    10 hours ago
















10














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




















  • 4




    Related stackoverflow.com/a/32424763/1746118
    – nullpointer
    10 hours ago














10












10








10


3





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









Peter Mortensen

13.5k1983111




13.5k1983111










asked 14 hours ago









fastcodejava

23.9k19109162




23.9k19109162








  • 4




    Related stackoverflow.com/a/32424763/1746118
    – nullpointer
    10 hours ago














  • 4




    Related stackoverflow.com/a/32424763/1746118
    – nullpointer
    10 hours ago








4




4




Related stackoverflow.com/a/32424763/1746118
– nullpointer
10 hours ago




Related stackoverflow.com/a/32424763/1746118
– nullpointer
10 hours ago












2 Answers
2






active

oldest

votes


















15














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. Along with the boxing iterate generates an infinite stream which in this specific case has more overhead than the finite one with rangeClosed. Further, it's far easier to run IntStream.rangeClosed in parallel, not that you want to in this specific case but it's something to keep in mind. here is some discussion on Generators as sources by Brian Goetz.


etc...






share|improve this answer























  • "iterate generates an infinite stream which again has more overhead than a finite one." why's that?
    – Alexander
    6 hours ago










  • @Alexander Firstly, I will improve my wording as it may not be the best to remove ambiguity. Thanks. I have also included a reference to a blog which might be of interest.
    – Aomine
    4 hours ago





















12














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
    11 hours ago






  • 1




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






  • 1




    @fastcodejava I have edited my answer to try to explain.
    – GBlodgett
    10 hours ago






  • 2




    1 👏🏻 for awesome explanation @GBlodgett
    – Deadpool
    10 hours ago











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









15














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. Along with the boxing iterate generates an infinite stream which in this specific case has more overhead than the finite one with rangeClosed. Further, it's far easier to run IntStream.rangeClosed in parallel, not that you want to in this specific case but it's something to keep in mind. here is some discussion on Generators as sources by Brian Goetz.


etc...






share|improve this answer























  • "iterate generates an infinite stream which again has more overhead than a finite one." why's that?
    – Alexander
    6 hours ago










  • @Alexander Firstly, I will improve my wording as it may not be the best to remove ambiguity. Thanks. I have also included a reference to a blog which might be of interest.
    – Aomine
    4 hours ago


















15














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. Along with the boxing iterate generates an infinite stream which in this specific case has more overhead than the finite one with rangeClosed. Further, it's far easier to run IntStream.rangeClosed in parallel, not that you want to in this specific case but it's something to keep in mind. here is some discussion on Generators as sources by Brian Goetz.


etc...






share|improve this answer























  • "iterate generates an infinite stream which again has more overhead than a finite one." why's that?
    – Alexander
    6 hours ago










  • @Alexander Firstly, I will improve my wording as it may not be the best to remove ambiguity. Thanks. I have also included a reference to a blog which might be of interest.
    – Aomine
    4 hours ago
















15












15








15






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. Along with the boxing iterate generates an infinite stream which in this specific case has more overhead than the finite one with rangeClosed. Further, it's far easier to run IntStream.rangeClosed in parallel, not that you want to in this specific case but it's something to keep in mind. here is some discussion on Generators as sources by Brian Goetz.


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. Along with the boxing iterate generates an infinite stream which in this specific case has more overhead than the finite one with rangeClosed. Further, it's far easier to run IntStream.rangeClosed in parallel, not that you want to in this specific case but it's something to keep in mind. here is some discussion on Generators as sources by Brian Goetz.


etc...







share|improve this answer














share|improve this answer



share|improve this answer








edited 4 hours ago

























answered 14 hours ago









Aomine

39.2k73669




39.2k73669












  • "iterate generates an infinite stream which again has more overhead than a finite one." why's that?
    – Alexander
    6 hours ago










  • @Alexander Firstly, I will improve my wording as it may not be the best to remove ambiguity. Thanks. I have also included a reference to a blog which might be of interest.
    – Aomine
    4 hours ago




















  • "iterate generates an infinite stream which again has more overhead than a finite one." why's that?
    – Alexander
    6 hours ago










  • @Alexander Firstly, I will improve my wording as it may not be the best to remove ambiguity. Thanks. I have also included a reference to a blog which might be of interest.
    – Aomine
    4 hours ago


















"iterate generates an infinite stream which again has more overhead than a finite one." why's that?
– Alexander
6 hours ago




"iterate generates an infinite stream which again has more overhead than a finite one." why's that?
– Alexander
6 hours ago












@Alexander Firstly, I will improve my wording as it may not be the best to remove ambiguity. Thanks. I have also included a reference to a blog which might be of interest.
– Aomine
4 hours ago






@Alexander Firstly, I will improve my wording as it may not be the best to remove ambiguity. Thanks. I have also included a reference to a blog which might be of interest.
– Aomine
4 hours ago















12














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
    11 hours ago






  • 1




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






  • 1




    @fastcodejava I have edited my answer to try to explain.
    – GBlodgett
    10 hours ago






  • 2




    1 👏🏻 for awesome explanation @GBlodgett
    – Deadpool
    10 hours ago
















12














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
    11 hours ago






  • 1




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






  • 1




    @fastcodejava I have edited my answer to try to explain.
    – GBlodgett
    10 hours ago






  • 2




    1 👏🏻 for awesome explanation @GBlodgett
    – Deadpool
    10 hours ago














12












12








12






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 10 hours ago

























answered 14 hours ago









GBlodgett

9,15641633




9,15641633








  • 1




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






  • 1




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






  • 1




    @fastcodejava I have edited my answer to try to explain.
    – GBlodgett
    10 hours ago






  • 2




    1 👏🏻 for awesome explanation @GBlodgett
    – Deadpool
    10 hours ago














  • 1




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






  • 1




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






  • 1




    @fastcodejava I have edited my answer to try to explain.
    – GBlodgett
    10 hours ago






  • 2




    1 👏🏻 for awesome explanation @GBlodgett
    – Deadpool
    10 hours ago








1




1




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




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




1




1




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




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




1




1




@fastcodejava I have edited my answer to try to explain.
– GBlodgett
10 hours ago




@fastcodejava I have edited my answer to try to explain.
– GBlodgett
10 hours ago




2




2




1 👏🏻 for awesome explanation @GBlodgett
– Deadpool
10 hours ago




1 👏🏻 for awesome explanation @GBlodgett
– Deadpool
10 hours ago


















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