Need Help with Comparator question in Stream

[Thomas Roth]

Hi folks,

i need help with following situation:

List<String> list = Arrays.asList(
               "Bob", "Sarah", "Marcus","Leander","Thomas");


I just can not understand why this works:

list.stream().sorted(  Comparator.comparing(String::length).thenComparing(s->s.charAt(2)) ).forEach(System.out::println);

while this does NOT work:

list.stream().sorted(  Comparator.comparing(s->s.length()).thenComparing(s->s.charAt(2)) ).forEach(System.out::println);

I know it must have something to do with the method references, but i just cant figure it out.

Similar problem, why does this then work:

Comparator<String> com1 = Comparator.comparing(s->s.length());
Comparator<String> com2 = com1.thenComparing(s->s.charAt(5));
list.stream().sorted( com1.thenComparing(s->s.charAt(2)) ).forEach(System.out::println);

but if i write for com1 the "Comparator.comparing(s->s.length())" in the sorted() it again wont work. ?

Any help would be appreciated

[Stephan van Hulst]

The way Java performs type inference on lambda expressions is very complex and will take a long time to explain. The short answer is that in the first case, you explicitly told the compiler to use the length() method of the String class, while in the second case, the compiler has to infer that you want to use the String class. Part of the reason why the compiler is not able to infer that you want to use String, is because comparing() takes a generic type with a parameter that has a lower bound. So when you write Comparator.comparing(s -> s.length()), the compiler interprets that as Comparator.comparing((Object s) -> s.length()). As you know, Object doesn't have a length() method. The compiler does not look at the surrounding method call to provide the comparing() method with more context.

You can fix this by explicitly stating the type of the lambda argument, or by passing the generic type arguments to the comparing() method explicitly:

Comparator.comparing((String s) -> s.length())

Comparator.<String, Integer> comparing(s -> s.length())

[Thomas Roth]

Stephan van Hulst wrote:The way Java performs type inference on lambda expressions is very complex and will take a long time to explain. The short answer is that in the first case, you explicitly told the compiler to use the length() method of the String class, while in the second case, the compiler has to infer that you want to use the String class. Part of the reason why the compiler is not able to infer that you want to use String, is because comparing() takes a generic type with a parameter that has a lower bound. So when you write Comparator.comparing(s -> s.length()), the compiler interprets that as Comparator.comparing((Object s) -> s.length()). As you know, Object doesn't have a length() method. The compiler does not look at the surrounding method call to provide the comparing() method with more context.

You can fix this by explicitly stating the type of the lambda argument, or by passing the generic type arguments to the comparing() method explicitly:

Comparator.comparing((String s) -> s.length())

Comparator.<String, Integer> comparing(s -> s.length())

Thank you very much Stephan, that was a very good explanation and helped me to understand what is going on there. ;)