Why can't you write it like this?
Comparator<Item> sort = Comparator.comparing(Item::isOpen);
Underneath Boolean.compareTo is called, which in turn is the same as Boolean.compare
public static int compare(boolean x, boolean y) {
return (x == y) ? 0 : (x ? 1 : -1);
}
And this: Comparator.comparing returns int and not "Item". make little sense, Comparator.comparing must return a Comparator<T>; in your case it correctly returns a Comparator<Item>.
Why can't you write it like this?
Comparator<Item> sort = Comparator.comparing(Item::isOpen);
Underneath Boolean.compareTo is called, which in turn is the same as Boolean.compare
public static int compare(boolean x, boolean y) {
return (x == y) ? 0 : (x ? 1 : -1);
}
And this: Comparator.comparing returns int and not "Item". make little sense, Comparator.comparing must return a Comparator<T>; in your case it correctly returns a Comparator<Item>.
The overloads comparingInt, comparingLong, and comparingDouble exist for performance reasons only. They are semantically identical to the unspecialized comparing method, so using comparing instead of comparingXXX has the same outcome, but might having boxing overhead, but the actual implications depend on the particular execution environment.
In case of boolean values, we can predict that the overhead will be negligible, as the method Boolean.valueOf will always return either Boolean.TRUE or Boolean.FALSE and never create new instances, so even if a particular JVM fails to inline the entire code, it does not depend on the presence of Escape Analysis in the optimizer.
As you already figured out, reversing a comparator is implemented by swapping the argument internally, just like you did manually in your lambda expression.
Note that it is still possible to create a comparator fusing the reversal and an unboxed comparison without having to repeat the isOpen() expression:
Comparator<Item> sort = Comparator.comparingInt(i -> i.isOpen()? 0: 1);
but, as said, it’s unlikely to have a significantly higher performance than the Comparator.comparing(Item::isOpen).reversed() approach.
But note that if you have a boolean sort criteria and care for the maximum performance, you may consider replacing the general-purpose sort algorithm with a bucket sort variant. E.g.
If you have a Stream, replace
List<Item> result = /* stream of Item */
.sorted(Comparator.comparing(Item::isOpen).reversed())
.collect(Collectors.toList());
with
Map<Boolean,List<Item>> map = /* stream of Item */
.collect(Collectors.partitioningBy(Item::isOpen,
Collectors.toCollection(ArrayList::new)));
List<Item> result = map.get(true);
result.addAll(map.get(false));
or, if you have a List, replace
list.sort(Comparator.comparing(Item::isOpen).reversed());
with
ArrayList<Item> temp = new ArrayList<>(list.size());
list.removeIf(item -> !item.isOpen() && temp.add(item));
list.addAll(temp);
etc.
You can look up how it is implemented for the java.lang.Boolean, since that class, naturally, uses a primitive boolean as well:
public int compareTo(Boolean b) {
return (b.value == value ? 0 : (value ? 1 : -1));
}
As of Java 7 you can simply use the built-in static method Boolean.compare(a, b).
Since Java 7, the logic that Marko Topolnik showed in his answer has moved into another method to expose a way to compare primitive boolean.
Javadoc for Boolean.compare(boolean x, boolean y):
public static int compare(boolean x, boolean y)
Compares two boolean values. The value returned is identical to
what would be returned by:
Boolean.valueOf(x).compareTo(Boolean.valueOf(y))