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

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











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












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










Your Answer






StackExchange.ifUsing("editor", function ()
StackExchange.using("externalEditor", function ()
StackExchange.using("snippets", function ()
StackExchange.snippets.init();
);
);
, "code-snippets");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "1"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);













 

draft saved


draft discarded


















StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53219523%2fwhy-is-the-stream-sorted-not-type-safe-in-java-8%23new-answer', 'question_page');

);

Post as a guest

















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

















 

draft saved


draft discarded















































 


draft saved


draft discarded














StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53219523%2fwhy-is-the-stream-sorted-not-type-safe-in-java-8%23new-answer', 'question_page');

);

Post as a guest
































































Comments

Post a Comment

Popular posts from this blog

Long meetings (6-7 hours a day): Being “babysat” by supervisor

Image
Clash Royale CLAN TAG #URR8PPP .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty margin-bottom:0; up vote 31 down vote favorite 4 I hold a PhD in mechanical engineering with a 7 year background in self-guided, self-paced research including 2 years as a graduate level course instructor and a subject matter expert for various engineering projects in a university setting in the United States. For the last 10-11 months, I am working in a research industry in France where my task is to debug C++ spaghetti code written over a period of several years by my current supervisor. It is a team of 2: just I and my supervisor. Diurnally, I am posed with a "bug of the day" to get out of this C++ code. My supervisor generally gives me about 15-30 minutes to solve it and then accosts me about whether or not the task is completed and then rinses and repeats this until the end of the day. Off-late, my supervisor has been holding 6-7 hour long "meetin
Read more

Is the Concept of Multiple Fantasy Races Scientifically Flawed? [closed]

Image
Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite Many fantasy worlds have multiple humanoid species (elves, orcs, humans, etc) living in the same world, in addition to that, they often have the ability to interbreed. And I can't help but question their believability... fantasy-races reproduction interspecies-relations share | improve this question edited Aug 9 at 14:46 Michael Kjörling ♦ 21.1k 10 85 161 asked Aug 9 at 13:13 Chlodio 118 6 closed as unclear what you're asking by MichaelK, Gryphon, John, Vincent, Frostfyre Aug 9 at 15:25 Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question. 3 Stack Exchange exist to
Read more

Confectionery

Image
Prepared foods made with a lot of sugar or other cabohydrates "Sweetmeat" redirects here. For the racehorse, see Sweetmeat (horse). Not to be confused with Sweetbread. This Kransekake is a traditional Scandinavian baker's confection. Confectionery is the art of making confections , which are food items that are rich in sugar and carbohydrates. Exact definitions are difficult. [1] In general, though, confectionery is divided into two broad and somewhat overlapping categories, bakers' confections and sugar confections. [2] Bakers' confectionery, also called flour confections , includes principally sweet pastries, cakes, and similar baked goods. Sugar confectionery includes candies ( sweets in British English), candied nuts, chocolates, chewing gum, bubble gum, pastillage, and other confections that are made primarily of sugar. In some cases, chocolate confections (confections made of chocolate) are treated as a separate category, as are sugar-free versions of
Read more