On Fri, 26 Nov 2021 14:31:04 GMT, Julia Boes <jb...@openjdk.org> wrote:
>> make/jdk/src/classes/build/tools/taglet/ToolGuide.java line 159:
>>
>>> 157: .toString()
>>> 158: .replace('.', '/')
>>> 159: .replaceAll("[^/]+", "..");
>>
>> If the class is in a module don't you have to get one step higher to get the
>> root?
>> I am not familiar with this code, so I'm just reasoning by induction here -
>> trying to match with what the case for PACKAGE seems to be doing...
>
> Same here, I initially applied the same pattern as for PACKAGE but that does
> not produce the correct path (it includes 1 ".." too much.):
>
> String typePart = te.getQualifiedName()
> .toString()
> .replace('.', '/')
> .replaceAll("[^/]+", "..");
> return te.getEnclosingElement().getEnclosingElement() != null
> ? "../" + typePart
> : typePart;
I see - class has a class name which is a file - so `foo.Bar` produces `../..`
relative to the directory foo (foo/../..) - so it already has an additional
`..` compared to package - where `a.b` would produce ../.. relative to
*directory* b which would lead to `b/../..` and not `a/../..` which is what we
need.
The difference comes from the fact that `Bar` is a file whereas `b` is a
directory.
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PR: https://git.openjdk.java.net/jdk/pull/6566
