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3070. path::lexically_relative causes surprising results if a filename can also be a root-name

Section: [fs.path.gen] Status: C++20 Submitter: Billy O'Neal III Opened: 2018-02-23 Last modified: 2021-02-25

Priority: 2

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path::lexically_relative constructs the resulting path with operator/=. If any of the filename elements from *this are themselves acceptable root-names, operator/= will destroy any previous value, and take that root_name(). For example:


On a POSIX implementation, this would return path("../b:"), but on a Windows implementation, the "b:" element is interpreted as a root-name, and clobbers the entire result path, giving path("b:"). We should detect this problematic condition and fail (by returning path()).

[2019-01-20 Reflector prioritization]

Set Priority to 2

[2019 Cologne Wednesday night]

Status to Ready

Proposed resolution:

This wording is relative to N4727.

  1. Change [fs.path.gen] as indicated:

    path lexically_relative(const path& base) const;

    -3- […]

    -4- Effects: If root_name() != base.root_name() is true or is_absolute() != base.is_absolute() is true or !has_root_directory() && base.has_root_directory() is true or if any filename in relative_path() or base.relative_path() can be interpreted as a root-name, returns path(). [Note: On a POSIX implementation, no filename in a relative-path is acceptable as a root-nameend note] Determines the first mismatched element of *this and base as if by:

    auto [a, b] = mismatch(begin(), end(), base.begin(), base.end());


    1. (4.1) — if a == end() and b == base.end(), returns path("."); otherwise

    2. (4.2) — let n be the number of filename elements in [b, base.end()) that are not dot or dot-dot minus the number that are dot-dot. If n < 0, returns path(); otherwise

    3. (4.3) — returns an object of class path that is default-constructed, followed by

      1. (4.3.1) — application of operator/=(path("..")) n times, and then

      2. (4.3.2) — application of operator/= for each element in [a, end()).