Choose pathlib lexical paths versus resolved paths in a local tool

Quick answer

Path spelling and resolved target answer different local-tool questions: one is what the caller typed, the other follows filesystem links.

Example

The temporary fixture makes a directory symlink. link.name preserves the lexical display name, while link.resolve() identifies target. The assertion demonstrates why a status message and a target-identity check may intentionally use different values.

from pathlib import Path
from tempfile import TemporaryDirectory
with TemporaryDirectory() as d:
    base = Path(d)
    target = base / 'target'
    target.mkdir()
    link = base / 'link'
    link.symlink_to(target, target_is_directory=True)
    assert str(link).endswith('link') and link.resolve() == target.resolve()
    print(f'lexical={link.name} resolved={link.resolve().name}')

Expected stdout:

lexical=link resolved=target

Reading the result

Resolution can fail or change meaning when links are broken or permissions differ. Do not resolve merely for prettier output; resolve only when the tool rule is about the target.

For a destructive operation, target identity may matter more than display spelling; for a diagnostic, the original argument is often kinder. Make the choice at each call site explicit.

The symlink lives inside TemporaryDirectory, so the comparison has no dependence on a user path. Its only purpose is to make the lexical-versus-target distinction observable.

Sources

- Python pathlib documentation

Prepared with AI assistance. The example uses synthetic data; its stated limits apply.

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