是否存在可能的身份表示Comparator
那可能存在吗?
在寻找简化代码的过程中删除Java中重载的方法 https://stackoverflow.com/questions/58782150/removing-overloaded-method-in-java/58782788#58782788,我考虑了这一点并最终得出结论,如果每次比较的结果都是相等的,那么顺序不会真正改变,从而使操作成为identity。因此我最终得到了(一个低效的)建议,如下所示:
public static <T, G> List<G> toListOfNewType(List<T> inputList, Function<T, G> mapperFunction) {
return toListOfNewType(inputList, mapperFunction, (a, b) -> 0); // overloaded with comparator for 'G' type
}
但我现在想知道的是,这是否适用于具有其自定义的对象compareTo
实施也是如此?考虑到这一点真的安全吗?Stream
执行?
Edit:我尝试过并保留顺序的某些测试如下:
List<Integer> integers = List.of(1, 3, 45356, 47424, 34234, 4, 4, 234234, 234, 0, -23, -34);
System.out.println(integers);
System.out.println(integers.stream().sorted((a, b) -> 0).collect(Collectors.toList()));
System.out.println(integers.stream().sorted((a, b) -> 0).parallel().collect(Collectors.toList()));
[1, 3, 45356, 47424, 34234, 4, 4, 234234, 234, 0, -23, -34]
[1, 3, 45356, 47424, 34234, 4, 4, 234234, 234, 0, -23, -34]
[1, 3, 45356, 47424, 34234, 4, 4, 234234, 234, 0, -23, -34]
List<String> strings = List.of("aadad", "Z", "vsadasd", "zadad", "C", "Aadasd");
System.out.println(strings);
System.out.println(strings.stream().sorted((a, b) -> 0).collect(Collectors.toList()));
System.out.println(strings.stream().sorted((a, b) -> 0).parallel().collect(Collectors.toList()));
[aadad, Z, vsadasd, zadad, C, Aadasd]
[aadad, Z, vsadasd, zadad, C, Aadasd]
[aadad, Z, vsadasd, zadad, C, Aadasd]
Set<Integer> integerSet = Set.of(1, 3, 45356, 47424, 34234, 4, 234234, 234, 0, -23, -34);
System.out.println(integerSet);
System.out.println(integerSet.stream().sorted((a, b) -> 0).parallel().collect(Collectors.toList()));
System.out.println(integerSet.stream().sorted((a, b) -> 0).collect(Collectors.toList()));
[-34, 45356, 47424, 234, -23, 234234, 1, 34234, 3, 4, 0]
[-34, 45356, 47424, 234, -23, 234234, 1, 34234, 3, 4, 0]
[-34, 45356, 47424, 234, -23, 234234, 1, 34234, 3, 4, 0]
Set<String> stringSet = Set.of("aadad", "Z", "vsadasd", "zadad", "C", "Aadasd");
System.out.println(stringSet);
System.out.println(stringSet.stream().sorted((a, b) -> 0).collect(Collectors.toList()));
System.out.println(stringSet.stream().sorted((a, b) -> 0).parallel().collect(Collectors.toList()));
[zadad, Z, vsadasd, C, Aadasd, aadad]
[zadad, Z, vsadasd, C, Aadasd, aadad]
[zadad, Z, vsadasd, C, Aadasd, aadad]