bpo-33254: do not return an empty list when asking for the contents of a…

bpo-33254: do not return an empty list when asking for the contents of a namespace package (GH-6467)

(cherry picked from commit 3ab9365d)
Co-authored-by: Brett Cannon <brettcannon@users.noreply.github.com>

Doc/library/importlib.rst

@@ -530,7 +530,7 @@ ABC hierarchy::
 
     .. abstractmethod:: contents()
 
-        Returns an :term:`iterator` of strings over the contents of
+        Returns an :term:`iterable` of strings over the contents of
         the package. Do note that it is not required that all names
         returned by the iterator be actual resources, e.g. it is
         acceptable to return names for which :meth:`is_resource` would
@@ -544,7 +544,7 @@ ABC hierarchy::
         the file system then those subdirectory names can be used
         directly.
 
-        The abstract method returns an iterator of no items.
+        The abstract method returns an iterable of no items.
 
 
 .. class:: ResourceLoader
@@ -926,9 +926,9 @@ The following functions are available.
 
 .. function:: contents(package)
 
-    Return an iterator over the named items within the package.  The iterator
+    Return an iterable over the named items within the package.  The iterable
     returns :class:`str` resources (e.g. files) and non-resources
-    (e.g. directories).  The iterator does not recurse into subdirectories.
+    (e.g. directories).  The iterable does not recurse into subdirectories.
 
     *package* is either a name or a module object which conforms to the
     ``Package`` requirements.

Lib/importlib/abc.py

@@ -381,8 +381,8 @@ class ResourceReader(metaclass=abc.ABCMeta):
 
     @abc.abstractmethod
     def contents(self):
-        """Return an iterator of strings over the contents of the package."""
-        return iter([])
+        """Return an iterable of strings over the contents of the package."""
+        return []
 
 
 _register(ResourceReader, machinery.SourceFileLoader)

Lib/importlib/resources.py

@@ -9,7 +9,7 @@ from importlib.abc import ResourceLoader
 from io import BytesIO, TextIOWrapper
 from pathlib import Path
 from types import ModuleType
-from typing import Iterator, Optional, Set, Union   # noqa: F401
+from typing import Iterable, Iterator, Optional, Set, Union   # noqa: F401
 from typing import cast
 from typing.io import BinaryIO, TextIO
 from zipimport import ZipImportError
@@ -44,8 +44,7 @@ def _normalize_path(path) -> str:
 
     If the resulting string contains path separators, an exception is raised.
     """
-    str_path = str(path)
-    parent, file_name = os.path.split(str_path)
+    parent, file_name = os.path.split(path)
     if parent:
         raise ValueError('{!r} must be only a file name'.format(path))
     else:
@@ -228,8 +227,8 @@ def is_resource(package: Package, name: str) -> bool:
     return path.is_file()
 
 
-def contents(package: Package) -> Iterator[str]:
-    """Return the list of entries in 'package'.
+def contents(package: Package) -> Iterable[str]:
+    """Return an iterable of entries in 'package'.
 
     Note that not all entries are resources.  Specifically, directories are
     not considered resources.  Use `is_resource()` on each entry returned here
@@ -238,15 +237,15 @@ def contents(package: Package) -> Iterator[str]:
     package = _get_package(package)
     reader = _get_resource_reader(package)
     if reader is not None:
-        yield from reader.contents()
-        return
+        return reader.contents()
     # Is the package a namespace package?  By definition, namespace packages
     # cannot have resources.  We could use _check_location() and catch the
     # exception, but that's extra work, so just inline the check.
-    if package.__spec__.origin is None or not package.__spec__.has_location:
-        return []
-    package_directory = Path(package.__spec__.origin).parent
-    yield from os.listdir(str(package_directory))
+    elif package.__spec__.origin is None or not package.__spec__.has_location:
+        return ()
+    else:
+        package_directory = Path(package.__spec__.origin).parent
+        return os.listdir(package_directory)
 
 
 # Private implementation of ResourceReader and get_resource_reader() for

Lib/test/test_importlib/test_resource.py

@@ -131,10 +131,9 @@ class ResourceFromZipsTest(util.ZipSetupBase, unittest.TestCase):
 
 
 class NamespaceTest(unittest.TestCase):
-    def test_namespaces_cant_have_resources(self):
-        contents = set(resources.contents(
-            'test.test_importlib.data03.namespace'))
-        self.assertEqual(len(contents), 0)
+    def test_namespaces_cannot_have_resources(self):
+        contents = resources.contents('test.test_importlib.data03.namespace')
+        self.assertFalse(list(contents))
         # Even though there is a file in the namespace directory, it is not
         # considered a resource, since namespace packages can't have them.
         self.assertFalse(resources.is_resource(

Misc/NEWS.d/next/Library/2018-04-13-15-14-47.bpo-33254.DS4KFK.rst

+Have :func:`importlib.resources.contents` and
+:meth:`importlib.abc.ResourceReader.contents` return an :term:`iterable` instead
+of an :term:`iterator`.