Why is the Stream.sorted not type safe in Java 8?

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up vote
8
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favorite












This is from the Stream interface from Oracles implementation of JDK 8:



public interface Stream<T> extends BaseStream<T, Stream<T>> 
 Stream<T> sorted();


and it is very easy to blow this in run-time and no warning will be generated in compile time, here is an example:



class Foo 
 public static void main(String args) 
 Arrays.asList(new Foo(), new Foo()).stream().sorted().forEach(f -> );



which will compile just fine but will throw an exception in runtime:



Exception in thread "main" java.lang.ClassCastException: Foo cannot be cast to java.lang.Comparable


What can be the reason that sorted method was not defined where compiler could actually catch such problems? Maybe I am wrong but isn 't it this simple:



interface Stream<T> 
 <C extends Comparable<T>> void sorted(C c);



?



Obviously the guys implementing this (who are light years ahead of me as far as programming and engineering is considered) must have a very good reason that I am unable to see, but what is that reason?






java java-8 java-stream comparable







share|improve this question



















  • 1




    You mean Comparator? There's already an overload for that. Not sure why you need C though.
    – shmosel
    1 hour ago







  • 2




    It's not possible to do that.
    – shmosel
    1 hour ago






  • 1




    Due to the nature of generics. The whole point of generics is that you're implementing generic functionality, which precludes type-specific logic. Your sample code doesn't make any sense. I think you're trying to redefine T (which isn't possible), but you're actually just defining a type for a useless argument.
    – shmosel
    1 hour ago







  • 1




    In your sorted example you're requiring an argument of type ? extends Comparable. Where is this argument going to come from?
    – Slaw
    1 hour ago







  • 1




    but sorted does not take any arguments as input - where are you going to take this argument from?
    – Eugene
    1 hour ago















up vote
8
down vote

favorite












This is from the Stream interface from Oracles implementation of JDK 8:



public interface Stream<T> extends BaseStream<T, Stream<T>> 
 Stream<T> sorted();


and it is very easy to blow this in run-time and no warning will be generated in compile time, here is an example:



class Foo 
 public static void main(String args) 
 Arrays.asList(new Foo(), new Foo()).stream().sorted().forEach(f -> );



which will compile just fine but will throw an exception in runtime:



Exception in thread "main" java.lang.ClassCastException: Foo cannot be cast to java.lang.Comparable


What can be the reason that sorted method was not defined where compiler could actually catch such problems? Maybe I am wrong but isn 't it this simple:



interface Stream<T> 
 <C extends Comparable<T>> void sorted(C c);



?



Obviously the guys implementing this (who are light years ahead of me as far as programming and engineering is considered) must have a very good reason that I am unable to see, but what is that reason?






java java-8 java-stream comparable







share|improve this question



















  • 1




    You mean Comparator? There's already an overload for that. Not sure why you need C though.
    – shmosel
    1 hour ago







  • 2




    It's not possible to do that.
    – shmosel
    1 hour ago






  • 1




    Due to the nature of generics. The whole point of generics is that you're implementing generic functionality, which precludes type-specific logic. Your sample code doesn't make any sense. I think you're trying to redefine T (which isn't possible), but you're actually just defining a type for a useless argument.
    – shmosel
    1 hour ago







  • 1




    In your sorted example you're requiring an argument of type ? extends Comparable. Where is this argument going to come from?
    – Slaw
    1 hour ago







  • 1




    but sorted does not take any arguments as input - where are you going to take this argument from?
    – Eugene
    1 hour ago













up vote
8
down vote

favorite









up vote
8
down vote

favorite











This is from the Stream interface from Oracles implementation of JDK 8:



public interface Stream<T> extends BaseStream<T, Stream<T>> 
 Stream<T> sorted();


and it is very easy to blow this in run-time and no warning will be generated in compile time, here is an example:



class Foo 
 public static void main(String args) 
 Arrays.asList(new Foo(), new Foo()).stream().sorted().forEach(f -> );



which will compile just fine but will throw an exception in runtime:



Exception in thread "main" java.lang.ClassCastException: Foo cannot be cast to java.lang.Comparable


What can be the reason that sorted method was not defined where compiler could actually catch such problems? Maybe I am wrong but isn 't it this simple:



interface Stream<T> 
 <C extends Comparable<T>> void sorted(C c);



?



Obviously the guys implementing this (who are light years ahead of me as far as programming and engineering is considered) must have a very good reason that I am unable to see, but what is that reason?






java java-8 java-stream comparable







share|improve this question















This is from the Stream interface from Oracles implementation of JDK 8:



public interface Stream<T> extends BaseStream<T, Stream<T>> 
 Stream<T> sorted();


and it is very easy to blow this in run-time and no warning will be generated in compile time, here is an example:



class Foo 
 public static void main(String args) 
 Arrays.asList(new Foo(), new Foo()).stream().sorted().forEach(f -> );



which will compile just fine but will throw an exception in runtime:



Exception in thread "main" java.lang.ClassCastException: Foo cannot be cast to java.lang.Comparable


What can be the reason that sorted method was not defined where compiler could actually catch such problems? Maybe I am wrong but isn 't it this simple:



interface Stream<T> 
 <C extends Comparable<T>> void sorted(C c);



?



Obviously the guys implementing this (who are light years ahead of me as far as programming and engineering is considered) must have a very good reason that I am unable to see, but what is that reason?






java java-8 java-stream comparable




java java-8 java-stream comparable






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 1 hour ago

























asked 1 hour ago









Koray Tugay

8,11926108210




8,11926108210







  • 1




    You mean Comparator? There's already an overload for that. Not sure why you need C though.
    – shmosel
    1 hour ago







  • 2




    It's not possible to do that.
    – shmosel
    1 hour ago






  • 1




    Due to the nature of generics. The whole point of generics is that you're implementing generic functionality, which precludes type-specific logic. Your sample code doesn't make any sense. I think you're trying to redefine T (which isn't possible), but you're actually just defining a type for a useless argument.
    – shmosel
    1 hour ago







  • 1




    In your sorted example you're requiring an argument of type ? extends Comparable. Where is this argument going to come from?
    – Slaw
    1 hour ago







  • 1




    but sorted does not take any arguments as input - where are you going to take this argument from?
    – Eugene
    1 hour ago













  • 1




    You mean Comparator? There's already an overload for that. Not sure why you need C though.
    – shmosel
    1 hour ago







  • 2




    It's not possible to do that.
    – shmosel
    1 hour ago






  • 1




    Due to the nature of generics. The whole point of generics is that you're implementing generic functionality, which precludes type-specific logic. Your sample code doesn't make any sense. I think you're trying to redefine T (which isn't possible), but you're actually just defining a type for a useless argument.
    – shmosel
    1 hour ago







  • 1




    In your sorted example you're requiring an argument of type ? extends Comparable. Where is this argument going to come from?
    – Slaw
    1 hour ago







  • 1




    but sorted does not take any arguments as input - where are you going to take this argument from?
    – Eugene
    1 hour ago








1




1




You mean Comparator? There's already an overload for that. Not sure why you need C though.
– shmosel
1 hour ago





You mean Comparator? There's already an overload for that. Not sure why you need C though.
– shmosel
1 hour ago





2




2




It's not possible to do that.
– shmosel
1 hour ago




It's not possible to do that.
– shmosel
1 hour ago




1




1




Due to the nature of generics. The whole point of generics is that you're implementing generic functionality, which precludes type-specific logic. Your sample code doesn't make any sense. I think you're trying to redefine T (which isn't possible), but you're actually just defining a type for a useless argument.
– shmosel
1 hour ago





Due to the nature of generics. The whole point of generics is that you're implementing generic functionality, which precludes type-specific logic. Your sample code doesn't make any sense. I think you're trying to redefine T (which isn't possible), but you're actually just defining a type for a useless argument.
– shmosel
1 hour ago





1




1




In your sorted example you're requiring an argument of type ? extends Comparable. Where is this argument going to come from?
– Slaw
1 hour ago





In your sorted example you're requiring an argument of type ? extends Comparable. Where is this argument going to come from?
– Slaw
1 hour ago





1




1




but sorted does not take any arguments as input - where are you going to take this argument from?
– Eugene
1 hour ago





but sorted does not take any arguments as input - where are you going to take this argument from?
– Eugene
1 hour ago













2 Answers
2






active

oldest

votes

















up vote
3
down vote



accepted










How would you implement that? sorted is a intermediate operation (can be called anywhere between other intermediate operations), meaning you can start with a stream that is not Comparable, but call sorted on one that is Comparable:



Arrays.asList(new Foo(), new Foo())
 .stream()
 .map(Foo::getName) // name is a String for example
 .sorted()
 .forEach(f -> );


The thing that you are proposing takes an argument as input, but Stream::sorted does not, so you can't do that. The overload version accepts a Comparator - meaning you can sort something by a property, but still return Stream<T>. I think that this is quite easy to understand if you would try to write your minimal skeleton of a Stream interface/implementation.






share|improve this answer


















  • 1




    Well sorted in this example knows the type again, that is a Stream<String>, isn 't it?
    – Koray Tugay
    1 hour ago










  • @KorayTugay well yes, what is your point?
    – Eugene
    1 hour ago

















up vote
5
down vote













The documentation for Stream#sorted explains it perfectly:




Returns a stream consisting of the elements of this stream, sorted according to natural order. If the elements of this stream are not Comparable, a java.lang.ClassCastException may be thrown when the terminal operation is executed.




You're using the overloaded method that accepts no arguments (not the one that accepts a Comparator), and Foo does not implement Comparable.



If you're asking why the method doesn't throw a compiler error if the contents of the Stream do not implement Comparable, it would be because T is not forced to extend Comparable, and T cannot be changed without a call to Stream#map; it seems to only be a convenience method so no explicit Comparator needs to be provided when the elements already implement Comparable.



For it to be type-safe, T would have to extend Comparable, but that would be ridiculous, as it would prevent a stream from containing any objects that aren't Comparable.






share|improve this answer






















  • it would be because T is not forced to extend Comparable I know, but why does sorted accept a T but not something like what I suggest at the end of my question?
    – Koray Tugay
    1 hour ago










  • There's already an overloaded Stream#sorted method that accepts a Comparator. It accepting a Comparable wouldn't make much sense.
    – Jacob G.
    1 hour ago










  • I guess the point you are trying to make is that since sorted is part of the Stream class (which has type parameter T) T should not implement Comparable as that will force all types using the stream pipeline to implement Comparable.
    – user7
    1 hour ago










  • @user7 Exactly :) which is why the documentation states "sorted according to natural order". Objects that don't implement Comparable don't have a natural order.
    – Jacob G.
    1 hour ago










  • @user7 Yes, obviously that would be a disaster.. But can we not just define a method parameters type to be of generic type of the interface AND restrict it to also something else? (That is what I tried in the last code snipped in my question, at least I tried to..)
    – Koray Tugay
    1 hour ago










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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
3
down vote



accepted










How would you implement that? sorted is a intermediate operation (can be called anywhere between other intermediate operations), meaning you can start with a stream that is not Comparable, but call sorted on one that is Comparable:



Arrays.asList(new Foo(), new Foo())
 .stream()
 .map(Foo::getName) // name is a String for example
 .sorted()
 .forEach(f -> );


The thing that you are proposing takes an argument as input, but Stream::sorted does not, so you can't do that. The overload version accepts a Comparator - meaning you can sort something by a property, but still return Stream<T>. I think that this is quite easy to understand if you would try to write your minimal skeleton of a Stream interface/implementation.






share|improve this answer


















  • 1




    Well sorted in this example knows the type again, that is a Stream<String>, isn 't it?
    – Koray Tugay
    1 hour ago










  • @KorayTugay well yes, what is your point?
    – Eugene
    1 hour ago














up vote
3
down vote



accepted










How would you implement that? sorted is a intermediate operation (can be called anywhere between other intermediate operations), meaning you can start with a stream that is not Comparable, but call sorted on one that is Comparable:



Arrays.asList(new Foo(), new Foo())
 .stream()
 .map(Foo::getName) // name is a String for example
 .sorted()
 .forEach(f -> );


The thing that you are proposing takes an argument as input, but Stream::sorted does not, so you can't do that. The overload version accepts a Comparator - meaning you can sort something by a property, but still return Stream<T>. I think that this is quite easy to understand if you would try to write your minimal skeleton of a Stream interface/implementation.






share|improve this answer


















  • 1




    Well sorted in this example knows the type again, that is a Stream<String>, isn 't it?
    – Koray Tugay
    1 hour ago










  • @KorayTugay well yes, what is your point?
    – Eugene
    1 hour ago












up vote
3
down vote



accepted







up vote
3
down vote



accepted






How would you implement that? sorted is a intermediate operation (can be called anywhere between other intermediate operations), meaning you can start with a stream that is not Comparable, but call sorted on one that is Comparable:



Arrays.asList(new Foo(), new Foo())
 .stream()
 .map(Foo::getName) // name is a String for example
 .sorted()
 .forEach(f -> );


The thing that you are proposing takes an argument as input, but Stream::sorted does not, so you can't do that. The overload version accepts a Comparator - meaning you can sort something by a property, but still return Stream<T>. I think that this is quite easy to understand if you would try to write your minimal skeleton of a Stream interface/implementation.






share|improve this answer














How would you implement that? sorted is a intermediate operation (can be called anywhere between other intermediate operations), meaning you can start with a stream that is not Comparable, but call sorted on one that is Comparable:



Arrays.asList(new Foo(), new Foo())
 .stream()
 .map(Foo::getName) // name is a String for example
 .sorted()
 .forEach(f -> );


The thing that you are proposing takes an argument as input, but Stream::sorted does not, so you can't do that. The overload version accepts a Comparator - meaning you can sort something by a property, but still return Stream<T>. I think that this is quite easy to understand if you would try to write your minimal skeleton of a Stream interface/implementation.







share|improve this answer














share|improve this answer



share|improve this answer








edited 1 hour ago

























answered 1 hour ago









Eugene

65.8k992156




65.8k992156







  • 1




    Well sorted in this example knows the type again, that is a Stream<String>, isn 't it?
    – Koray Tugay
    1 hour ago










  • @KorayTugay well yes, what is your point?
    – Eugene
    1 hour ago












  • 1




    Well sorted in this example knows the type again, that is a Stream<String>, isn 't it?
    – Koray Tugay
    1 hour ago










  • @KorayTugay well yes, what is your point?
    – Eugene
    1 hour ago







1




1




Well sorted in this example knows the type again, that is a Stream<String>, isn 't it?
– Koray Tugay
1 hour ago




Well sorted in this example knows the type again, that is a Stream<String>, isn 't it?
– Koray Tugay
1 hour ago












@KorayTugay well yes, what is your point?
– Eugene
1 hour ago




@KorayTugay well yes, what is your point?
– Eugene
1 hour ago












up vote
5
down vote













The documentation for Stream#sorted explains it perfectly:




Returns a stream consisting of the elements of this stream, sorted according to natural order. If the elements of this stream are not Comparable, a java.lang.ClassCastException may be thrown when the terminal operation is executed.




You're using the overloaded method that accepts no arguments (not the one that accepts a Comparator), and Foo does not implement Comparable.



If you're asking why the method doesn't throw a compiler error if the contents of the Stream do not implement Comparable, it would be because T is not forced to extend Comparable, and T cannot be changed without a call to Stream#map; it seems to only be a convenience method so no explicit Comparator needs to be provided when the elements already implement Comparable.



For it to be type-safe, T would have to extend Comparable, but that would be ridiculous, as it would prevent a stream from containing any objects that aren't Comparable.






share|improve this answer






















  • it would be because T is not forced to extend Comparable I know, but why does sorted accept a T but not something like what I suggest at the end of my question?
    – Koray Tugay
    1 hour ago










  • There's already an overloaded Stream#sorted method that accepts a Comparator. It accepting a Comparable wouldn't make much sense.
    – Jacob G.
    1 hour ago










  • I guess the point you are trying to make is that since sorted is part of the Stream class (which has type parameter T) T should not implement Comparable as that will force all types using the stream pipeline to implement Comparable.
    – user7
    1 hour ago










  • @user7 Exactly :) which is why the documentation states "sorted according to natural order". Objects that don't implement Comparable don't have a natural order.
    – Jacob G.
    1 hour ago










  • @user7 Yes, obviously that would be a disaster.. But can we not just define a method parameters type to be of generic type of the interface AND restrict it to also something else? (That is what I tried in the last code snipped in my question, at least I tried to..)
    – Koray Tugay
    1 hour ago














up vote
5
down vote













The documentation for Stream#sorted explains it perfectly:




Returns a stream consisting of the elements of this stream, sorted according to natural order. If the elements of this stream are not Comparable, a java.lang.ClassCastException may be thrown when the terminal operation is executed.




You're using the overloaded method that accepts no arguments (not the one that accepts a Comparator), and Foo does not implement Comparable.



If you're asking why the method doesn't throw a compiler error if the contents of the Stream do not implement Comparable, it would be because T is not forced to extend Comparable, and T cannot be changed without a call to Stream#map; it seems to only be a convenience method so no explicit Comparator needs to be provided when the elements already implement Comparable.



For it to be type-safe, T would have to extend Comparable, but that would be ridiculous, as it would prevent a stream from containing any objects that aren't Comparable.






share|improve this answer






















  • it would be because T is not forced to extend Comparable I know, but why does sorted accept a T but not something like what I suggest at the end of my question?
    – Koray Tugay
    1 hour ago










  • There's already an overloaded Stream#sorted method that accepts a Comparator. It accepting a Comparable wouldn't make much sense.
    – Jacob G.
    1 hour ago










  • I guess the point you are trying to make is that since sorted is part of the Stream class (which has type parameter T) T should not implement Comparable as that will force all types using the stream pipeline to implement Comparable.
    – user7
    1 hour ago










  • @user7 Exactly :) which is why the documentation states "sorted according to natural order". Objects that don't implement Comparable don't have a natural order.
    – Jacob G.
    1 hour ago










  • @user7 Yes, obviously that would be a disaster.. But can we not just define a method parameters type to be of generic type of the interface AND restrict it to also something else? (That is what I tried in the last code snipped in my question, at least I tried to..)
    – Koray Tugay
    1 hour ago












up vote
5
down vote










up vote
5
down vote









The documentation for Stream#sorted explains it perfectly:




Returns a stream consisting of the elements of this stream, sorted according to natural order. If the elements of this stream are not Comparable, a java.lang.ClassCastException may be thrown when the terminal operation is executed.




You're using the overloaded method that accepts no arguments (not the one that accepts a Comparator), and Foo does not implement Comparable.



If you're asking why the method doesn't throw a compiler error if the contents of the Stream do not implement Comparable, it would be because T is not forced to extend Comparable, and T cannot be changed without a call to Stream#map; it seems to only be a convenience method so no explicit Comparator needs to be provided when the elements already implement Comparable.



For it to be type-safe, T would have to extend Comparable, but that would be ridiculous, as it would prevent a stream from containing any objects that aren't Comparable.






share|improve this answer














The documentation for Stream#sorted explains it perfectly:




Returns a stream consisting of the elements of this stream, sorted according to natural order. If the elements of this stream are not Comparable, a java.lang.ClassCastException may be thrown when the terminal operation is executed.




You're using the overloaded method that accepts no arguments (not the one that accepts a Comparator), and Foo does not implement Comparable.



If you're asking why the method doesn't throw a compiler error if the contents of the Stream do not implement Comparable, it would be because T is not forced to extend Comparable, and T cannot be changed without a call to Stream#map; it seems to only be a convenience method so no explicit Comparator needs to be provided when the elements already implement Comparable.



For it to be type-safe, T would have to extend Comparable, but that would be ridiculous, as it would prevent a stream from containing any objects that aren't Comparable.







share|improve this answer














share|improve this answer



share|improve this answer








edited 1 hour ago

























answered 1 hour ago









Jacob G.

13.8k41859




13.8k41859











  • it would be because T is not forced to extend Comparable I know, but why does sorted accept a T but not something like what I suggest at the end of my question?
    – Koray Tugay
    1 hour ago










  • There's already an overloaded Stream#sorted method that accepts a Comparator. It accepting a Comparable wouldn't make much sense.
    – Jacob G.
    1 hour ago










  • I guess the point you are trying to make is that since sorted is part of the Stream class (which has type parameter T) T should not implement Comparable as that will force all types using the stream pipeline to implement Comparable.
    – user7
    1 hour ago










  • @user7 Exactly :) which is why the documentation states "sorted according to natural order". Objects that don't implement Comparable don't have a natural order.
    – Jacob G.
    1 hour ago










  • @user7 Yes, obviously that would be a disaster.. But can we not just define a method parameters type to be of generic type of the interface AND restrict it to also something else? (That is what I tried in the last code snipped in my question, at least I tried to..)
    – Koray Tugay
    1 hour ago
















  • it would be because T is not forced to extend Comparable I know, but why does sorted accept a T but not something like what I suggest at the end of my question?
    – Koray Tugay
    1 hour ago










  • There's already an overloaded Stream#sorted method that accepts a Comparator. It accepting a Comparable wouldn't make much sense.
    – Jacob G.
    1 hour ago










  • I guess the point you are trying to make is that since sorted is part of the Stream class (which has type parameter T) T should not implement Comparable as that will force all types using the stream pipeline to implement Comparable.
    – user7
    1 hour ago










  • @user7 Exactly :) which is why the documentation states "sorted according to natural order". Objects that don't implement Comparable don't have a natural order.
    – Jacob G.
    1 hour ago










  • @user7 Yes, obviously that would be a disaster.. But can we not just define a method parameters type to be of generic type of the interface AND restrict it to also something else? (That is what I tried in the last code snipped in my question, at least I tried to..)
    – Koray Tugay
    1 hour ago















it would be because T is not forced to extend Comparable I know, but why does sorted accept a T but not something like what I suggest at the end of my question?
– Koray Tugay
1 hour ago




it would be because T is not forced to extend Comparable I know, but why does sorted accept a T but not something like what I suggest at the end of my question?
– Koray Tugay
1 hour ago












There's already an overloaded Stream#sorted method that accepts a Comparator. It accepting a Comparable wouldn't make much sense.
– Jacob G.
1 hour ago




There's already an overloaded Stream#sorted method that accepts a Comparator. It accepting a Comparable wouldn't make much sense.
– Jacob G.
1 hour ago












I guess the point you are trying to make is that since sorted is part of the Stream class (which has type parameter T) T should not implement Comparable as that will force all types using the stream pipeline to implement Comparable.
– user7
1 hour ago




I guess the point you are trying to make is that since sorted is part of the Stream class (which has type parameter T) T should not implement Comparable as that will force all types using the stream pipeline to implement Comparable.
– user7
1 hour ago












@user7 Exactly :) which is why the documentation states "sorted according to natural order". Objects that don't implement Comparable don't have a natural order.
– Jacob G.
1 hour ago




@user7 Exactly :) which is why the documentation states "sorted according to natural order". Objects that don't implement Comparable don't have a natural order.
– Jacob G.
1 hour ago












@user7 Yes, obviously that would be a disaster.. But can we not just define a method parameters type to be of generic type of the interface AND restrict it to also something else? (That is what I tried in the last code snipped in my question, at least I tried to..)
– Koray Tugay
1 hour ago




@user7 Yes, obviously that would be a disaster.. But can we not just define a method parameters type to be of generic type of the interface AND restrict it to also something else? (That is what I tried in the last code snipped in my question, at least I tried to..)
– Koray Tugay
1 hour ago

















 

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