relativize(): implementation dependent if both Paths have a root component?

[T Vergilio]

The documentation for Path states that:

Path relativize(Path other)
Constructs a relative path between this path and a given path.
Relativization is the inverse of resolution. This method attempts to construct a relative path that when resolved against this path, yields a path that locates the same file as the given path. For example, on UNIX, if this path is "/a/b" and the given path is "/a/b/c/d" then the resulting relative path would be "c/d". Where this path and the given path do not have a root component, then a relative path can be constructed. A relative path cannot be constructed if only one of the paths have a root component. Where both paths have a root component then it is implementation dependent if a relative path can be constructed. If this path and the given path are equal then an empty path is returned.

Does the statement in bold mean I can't do this?

        Path absoluteOx = Paths.get("/home/dir");
        Path absoluteWindows = Paths.get("c:", "home", "java");
        System.out.println(absoluteWindows.relativize(absoluteOx)); //IllegalArgumentException: 'other' is different type of Path

Or does it mean it's implementation dependent whether I can do this?

        Path absoluteWindows1 = Paths.get("d:", "users");
        Path absoluteWindows2 = Paths.get("c:", "home", "java");
        System.out.println(absoluteWindows1.relativize(absoluteWindows2)); // IllegalArgumentException: 'other' has different root

What I'm really worried about is: is it implementation-dependent whether I can do this (I always assumed it wasn't)?

        Path absoluteWindows1 = Paths.get("c:", "users");
        Path absoluteWindows2 = Paths.get("c:", "home", "java");
        System.out.println(absoluteWindows1.relativize(absoluteWindows2)); //..\home\java

[Jeanne Boyarsky]

Good question. Note that there are two variants of "implementation dependent" at play here - one is the implementation of the JDK and one is what OS you are on.

The exam isn't going to try to trick you with edge cases here. You just need to remember to "simple" logic that the first two are undefined behavior and the third one (where they are both c:) works per the spec.

[T Vergilio]

Phew! Thanks for that, Jeanne. I was fine with the happy path, but then the API documentation really confused me. Good to know the exam is not going to go crazy on the edge cases!