Math.toIntExact inside lambda expression?












3















I'm learning about lambda expressions.
Given a list of names, I want to count the numbers of names that start with N.



I did that: 



final static List<String> friends = Arrays.asList("Brian", "Nate", "Neal", "Raju", "Sara", "Scott");



public static int countFriendsStartWithN() {

    return  Math.toIntExact(friends

            .stream()

            .filter(name -> name.startsWith("N"))

            .count());

}


The call to the count method returns a primitive long but I want an int.

I used Math.toIntExact to get the long value as int.



Is it possible to get the int value directly inside the lambda expression?






java lambda







share|improve this question

























  • What do you mean by lambda experssion? That looks alright to me.

    – SamzSakerz
    Nov 16 '18 at 8:10






  • 3





    If you know the long value is not too large, you can simply cast it to int.

    – Eran
    Nov 16 '18 at 8:11






  • 2





    @Eran and you know it's not too large, because it is necessarily no greater than the number of items in a collection, which is an int.

    – Andy Turner
    Nov 16 '18 at 8:14











  • I just want to know if is possible to do the int conversion inside the stream process.

    – Jesus Zavarce
    Nov 16 '18 at 8:16
















3















I'm learning about lambda expressions.
Given a list of names, I want to count the numbers of names that start with N.



I did that: 



final static List<String> friends = Arrays.asList("Brian", "Nate", "Neal", "Raju", "Sara", "Scott");



public static int countFriendsStartWithN() {

    return  Math.toIntExact(friends

            .stream()

            .filter(name -> name.startsWith("N"))

            .count());

}


The call to the count method returns a primitive long but I want an int.

I used Math.toIntExact to get the long value as int.



Is it possible to get the int value directly inside the lambda expression?






java lambda







share|improve this question

























  • What do you mean by lambda experssion? That looks alright to me.

    – SamzSakerz
    Nov 16 '18 at 8:10






  • 3





    If you know the long value is not too large, you can simply cast it to int.

    – Eran
    Nov 16 '18 at 8:11






  • 2





    @Eran and you know it's not too large, because it is necessarily no greater than the number of items in a collection, which is an int.

    – Andy Turner
    Nov 16 '18 at 8:14











  • I just want to know if is possible to do the int conversion inside the stream process.

    – Jesus Zavarce
    Nov 16 '18 at 8:16














3












3








3








I'm learning about lambda expressions.
Given a list of names, I want to count the numbers of names that start with N.



I did that: 



final static List<String> friends = Arrays.asList("Brian", "Nate", "Neal", "Raju", "Sara", "Scott");



public static int countFriendsStartWithN() {

    return  Math.toIntExact(friends

            .stream()

            .filter(name -> name.startsWith("N"))

            .count());

}


The call to the count method returns a primitive long but I want an int.

I used Math.toIntExact to get the long value as int.



Is it possible to get the int value directly inside the lambda expression?






java lambda







share|improve this question
















I'm learning about lambda expressions.
Given a list of names, I want to count the numbers of names that start with N.



I did that: 



final static List<String> friends = Arrays.asList("Brian", "Nate", "Neal", "Raju", "Sara", "Scott");



public static int countFriendsStartWithN() {

    return  Math.toIntExact(friends

            .stream()

            .filter(name -> name.startsWith("N"))

            .count());

}


The call to the count method returns a primitive long but I want an int.

I used Math.toIntExact to get the long value as int.



Is it possible to get the int value directly inside the lambda expression?






java lambda




java lambda






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 16 '18 at 8:38







Jesus Zavarce

















asked Nov 16 '18 at 8:09









Jesus ZavarceJesus Zavarce

1,069819




1,069819













  • What do you mean by lambda experssion? That looks alright to me.

    – SamzSakerz
    Nov 16 '18 at 8:10






  • 3





    If you know the long value is not too large, you can simply cast it to int.

    – Eran
    Nov 16 '18 at 8:11






  • 2





    @Eran and you know it's not too large, because it is necessarily no greater than the number of items in a collection, which is an int.

    – Andy Turner
    Nov 16 '18 at 8:14











  • I just want to know if is possible to do the int conversion inside the stream process.

    – Jesus Zavarce
    Nov 16 '18 at 8:16



















  • What do you mean by lambda experssion? That looks alright to me.

    – SamzSakerz
    Nov 16 '18 at 8:10






  • 3





    If you know the long value is not too large, you can simply cast it to int.

    – Eran
    Nov 16 '18 at 8:11






  • 2





    @Eran and you know it's not too large, because it is necessarily no greater than the number of items in a collection, which is an int.

    – Andy Turner
    Nov 16 '18 at 8:14











  • I just want to know if is possible to do the int conversion inside the stream process.

    – Jesus Zavarce
    Nov 16 '18 at 8:16

















What do you mean by lambda experssion? That looks alright to me.

– SamzSakerz
Nov 16 '18 at 8:10





What do you mean by lambda experssion? That looks alright to me.

– SamzSakerz
Nov 16 '18 at 8:10




3




3





If you know the long value is not too large, you can simply cast it to int.

– Eran
Nov 16 '18 at 8:11





If you know the long value is not too large, you can simply cast it to int.

– Eran
Nov 16 '18 at 8:11




2




2





@Eran and you know it's not too large, because it is necessarily no greater than the number of items in a collection, which is an int.

– Andy Turner
Nov 16 '18 at 8:14





@Eran and you know it's not too large, because it is necessarily no greater than the number of items in a collection, which is an int.

– Andy Turner
Nov 16 '18 at 8:14













I just want to know if is possible to do the int conversion inside the stream process.

– Jesus Zavarce
Nov 16 '18 at 8:16





I just want to know if is possible to do the int conversion inside the stream process.

– Jesus Zavarce
Nov 16 '18 at 8:16












5 Answers
5






active

oldest

votes


















6














No, it is not possible to fit your call to toIntExact into your chain of method calls, your stream pipeline. This is because count is a terminal operation and returns a primitive long on which no method call is possible. A terminal operation is an operation that ends the stream pipeline and produces a result (or a side effect).



So I believe the best thing you can do is to live with the code you already have. IMHO it’s fine.






share|improve this answer































    5














    Well, here's a somewhat silly way of calculating the count as an int without casting:



    public static int countFriendsStartWithN() {
    
    return friends.stream()
    
                  .filter(name -> name.startsWith("N"))
    
                  .mapToInt (s -> 1)
    
                  .sum();
    
}





    share|improve this answer































      4














      You can't do anything inside the lambda expression you currently have, since that's a Predicate: it returns a boolean. Math.toIntExact returns an int.



      You can do it without the Math.toIntExact (or a simple cast) like so:



      return /* create stream, filter */
      
    .mapToInt(a -> 1).sum();


      But this is likely to be slower than doing what you are doing at the moment.






      share|improve this answer































        3














        Yet another option that is not really better - it is possible to use a collector that applies a finisher:



        public static int countFriendsStartWithN() {
        
    return friends.stream()
        
                .filter(name -> name.startsWith("N"))
        
                .collect(Collectors.collectingAndThen(Collectors.counting(), Math::toIntExact));
        
}


        This may have an advantage if you need it frequenty - you could build a utility method returning this Collector to make it reusable.






        share|improve this answer

































          2














          Here's a way to do this with reduce



          public static int countFriendsStartWithN2() {
          
    return friends
          
            .stream()
          
            .filter(name -> name.startsWith("N"))
          
            .map(s -> 1)
          
            .reduce(0, Integer::sum);
          
}





          share|improve this answer























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            5 Answers
            5






            active

            oldest

            votes








            5 Answers
            5






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            6














            No, it is not possible to fit your call to toIntExact into your chain of method calls, your stream pipeline. This is because count is a terminal operation and returns a primitive long on which no method call is possible. A terminal operation is an operation that ends the stream pipeline and produces a result (or a side effect).



            So I believe the best thing you can do is to live with the code you already have. IMHO it’s fine.






            share|improve this answer




























              6














              No, it is not possible to fit your call to toIntExact into your chain of method calls, your stream pipeline. This is because count is a terminal operation and returns a primitive long on which no method call is possible. A terminal operation is an operation that ends the stream pipeline and produces a result (or a side effect).



              So I believe the best thing you can do is to live with the code you already have. IMHO it’s fine.






              share|improve this answer


























                6












                6








                6







                No, it is not possible to fit your call to toIntExact into your chain of method calls, your stream pipeline. This is because count is a terminal operation and returns a primitive long on which no method call is possible. A terminal operation is an operation that ends the stream pipeline and produces a result (or a side effect).



                So I believe the best thing you can do is to live with the code you already have. IMHO it’s fine.






                share|improve this answer













                No, it is not possible to fit your call to toIntExact into your chain of method calls, your stream pipeline. This is because count is a terminal operation and returns a primitive long on which no method call is possible. A terminal operation is an operation that ends the stream pipeline and produces a result (or a side effect).



                So I believe the best thing you can do is to live with the code you already have. IMHO it’s fine.







                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered Nov 16 '18 at 8:14









                Ole V.V.Ole V.V.

                31.8k74257




                31.8k74257

























                    5














                    Well, here's a somewhat silly way of calculating the count as an int without casting:



                    public static int countFriendsStartWithN() {
                    
    return friends.stream()
                    
                  .filter(name -> name.startsWith("N"))
                    
                  .mapToInt (s -> 1)
                    
                  .sum();
                    
}





                    share|improve this answer




























                      5














                      Well, here's a somewhat silly way of calculating the count as an int without casting:



                      public static int countFriendsStartWithN() {
                      
    return friends.stream()
                      
                  .filter(name -> name.startsWith("N"))
                      
                  .mapToInt (s -> 1)
                      
                  .sum();
                      
}





                      share|improve this answer


























                        5












                        5








                        5







                        Well, here's a somewhat silly way of calculating the count as an int without casting:



                        public static int countFriendsStartWithN() {
                        
    return friends.stream()
                        
                  .filter(name -> name.startsWith("N"))
                        
                  .mapToInt (s -> 1)
                        
                  .sum();
                        
}





                        share|improve this answer













                        Well, here's a somewhat silly way of calculating the count as an int without casting:



                        public static int countFriendsStartWithN() {
                        
    return friends.stream()
                        
                  .filter(name -> name.startsWith("N"))
                        
                  .mapToInt (s -> 1)
                        
                  .sum();
                        
}






                        share|improve this answer












                        share|improve this answer



                        share|improve this answer










                        answered Nov 16 '18 at 8:18









                        EranEran

                        290k37478563




                        290k37478563























                            4














                            You can't do anything inside the lambda expression you currently have, since that's a Predicate: it returns a boolean. Math.toIntExact returns an int.



                            You can do it without the Math.toIntExact (or a simple cast) like so:



                            return /* create stream, filter */
                            
    .mapToInt(a -> 1).sum();


                            But this is likely to be slower than doing what you are doing at the moment.






                            share|improve this answer




























                              4














                              You can't do anything inside the lambda expression you currently have, since that's a Predicate: it returns a boolean. Math.toIntExact returns an int.



                              You can do it without the Math.toIntExact (or a simple cast) like so:



                              return /* create stream, filter */
                              
    .mapToInt(a -> 1).sum();


                              But this is likely to be slower than doing what you are doing at the moment.






                              share|improve this answer


























                                4












                                4








                                4







                                You can't do anything inside the lambda expression you currently have, since that's a Predicate: it returns a boolean. Math.toIntExact returns an int.



                                You can do it without the Math.toIntExact (or a simple cast) like so:



                                return /* create stream, filter */
                                
    .mapToInt(a -> 1).sum();


                                But this is likely to be slower than doing what you are doing at the moment.






                                share|improve this answer













                                You can't do anything inside the lambda expression you currently have, since that's a Predicate: it returns a boolean. Math.toIntExact returns an int.



                                You can do it without the Math.toIntExact (or a simple cast) like so:



                                return /* create stream, filter */
                                
    .mapToInt(a -> 1).sum();


                                But this is likely to be slower than doing what you are doing at the moment.







                                share|improve this answer












                                share|improve this answer



                                share|improve this answer










                                answered Nov 16 '18 at 8:18









                                Andy TurnerAndy Turner

                                84k983143




                                84k983143























                                    3














                                    Yet another option that is not really better - it is possible to use a collector that applies a finisher:



                                    public static int countFriendsStartWithN() {
                                    
    return friends.stream()
                                    
                .filter(name -> name.startsWith("N"))
                                    
                .collect(Collectors.collectingAndThen(Collectors.counting(), Math::toIntExact));
                                    
}


                                    This may have an advantage if you need it frequenty - you could build a utility method returning this Collector to make it reusable.






                                    share|improve this answer






























                                      3














                                      Yet another option that is not really better - it is possible to use a collector that applies a finisher:



                                      public static int countFriendsStartWithN() {
                                      
    return friends.stream()
                                      
                .filter(name -> name.startsWith("N"))
                                      
                .collect(Collectors.collectingAndThen(Collectors.counting(), Math::toIntExact));
                                      
}


                                      This may have an advantage if you need it frequenty - you could build a utility method returning this Collector to make it reusable.






                                      share|improve this answer




























                                        3












                                        3








                                        3







                                        Yet another option that is not really better - it is possible to use a collector that applies a finisher:



                                        public static int countFriendsStartWithN() {
                                        
    return friends.stream()
                                        
                .filter(name -> name.startsWith("N"))
                                        
                .collect(Collectors.collectingAndThen(Collectors.counting(), Math::toIntExact));
                                        
}


                                        This may have an advantage if you need it frequenty - you could build a utility method returning this Collector to make it reusable.






                                        share|improve this answer















                                        Yet another option that is not really better - it is possible to use a collector that applies a finisher:



                                        public static int countFriendsStartWithN() {
                                        
    return friends.stream()
                                        
                .filter(name -> name.startsWith("N"))
                                        
                .collect(Collectors.collectingAndThen(Collectors.counting(), Math::toIntExact));
                                        
}


                                        This may have an advantage if you need it frequenty - you could build a utility method returning this Collector to make it reusable.







                                        share|improve this answer














                                        share|improve this answer



                                        share|improve this answer








                                        edited Nov 16 '18 at 9:01

























                                        answered Nov 16 '18 at 8:56









                                        HulkHulk

                                        3,42812142




                                        3,42812142























                                            2














                                            Here's a way to do this with reduce



                                            public static int countFriendsStartWithN2() {
                                            
    return friends
                                            
            .stream()
                                            
            .filter(name -> name.startsWith("N"))
                                            
            .map(s -> 1)
                                            
            .reduce(0, Integer::sum);
                                            
}





                                            share|improve this answer




























                                              2














                                              Here's a way to do this with reduce



                                              public static int countFriendsStartWithN2() {
                                              
    return friends
                                              
            .stream()
                                              
            .filter(name -> name.startsWith("N"))
                                              
            .map(s -> 1)
                                              
            .reduce(0, Integer::sum);
                                              
}





                                              share|improve this answer


























                                                2












                                                2








                                                2







                                                Here's a way to do this with reduce



                                                public static int countFriendsStartWithN2() {
                                                
    return friends
                                                
            .stream()
                                                
            .filter(name -> name.startsWith("N"))
                                                
            .map(s -> 1)
                                                
            .reduce(0, Integer::sum);
                                                
}





                                                share|improve this answer













                                                Here's a way to do this with reduce



                                                public static int countFriendsStartWithN2() {
                                                
    return friends
                                                
            .stream()
                                                
            .filter(name -> name.startsWith("N"))
                                                
            .map(s -> 1)
                                                
            .reduce(0, Integer::sum);
                                                
}






                                                share|improve this answer












                                                share|improve this answer



                                                share|improve this answer










                                                answered Nov 16 '18 at 8:25









                                                SamzSakerzSamzSakerz

                                                1,6691627




                                                1,6691627






























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