Fix preserving sorted order of a collection

[chanseokoh]

toSet() makes sorted() irrelevant. Looking at other code locations, toList() is really intended.

Other changes are only minor simplifications.

[codecov]

Codecov Report

Merging #845 (1000e0b) into master (6a9f329) will increase coverage by 0.00%.
The diff coverage is 22.22%.

@@           Coverage Diff           @@
##           master     #845   +/-   ##
=======================================
  Coverage   90.04%   90.05%           
=======================================
  Files         151      151           
  Lines       15361    15360    -1     
  Branches     1105     1101    -4     
=======================================
  Hits        13832    13832           
+ Misses       1214     1213    -1     
  Partials      315      315           

Impacted Files	Coverage Δ
...google/api/generator/gapic/protoparser/Parser.java	46.42% <22.22%> (+0.08%)	⬆️

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[meltsufin]

I guess it's more about just preserving order, not necessarily "sorted order".

[chanseokoh]

I guess it's more about just preserving order, not necessarily "sorted order".

Just so you know, the collection is sorted with .sorted() before .collect(). So, preserving the sorted order in that sense.

            .sorted((s1, s2) -> s2.name().compareTo(s1.name()))
            .collect(Collectors.toList()));

[meltsufin]

I guess it's more about just preserving order, not necessarily "sorted order".

Just so you know, the collection is sorted with .sorted() before .collect(). So, preserving the sorted order in that sense.

            .sorted((s1, s2) -> s2.name().compareTo(s1.name()))
            .collect(Collectors.toList()));

I saw that, but I was curious if that's actually needed since your PR description said that it's irrelevant.

[chanseokoh]

Ah, yeah, the comment on that part says sorting is intended and needed.

    // Use a list to ensure ordering for deterministic tests.
    outputMixinServices.addAll(
        outputMixinServiceSet.stream()
            .sorted((s1, s2) -> s2.name().compareTo(s1.name()))

[chanseokoh]

Umm... actually whether sorting is necessary is unclear if it's only for deterministic order, but I think it's still a good idea to sort. Eliminates any concern about which implementations of a set is being used.