libstdc++: Define _versioned_namespace in xmethods.py

flake8 pointed out that is_specialization_of in xmethods.py looks at a
global that wasn't added to the file.  This patch correct the
oversight.

libstdc++-v3/ChangeLog:

	* python/libstdcxx/v6/xmethods.py (_versioned_namespace):
	Define.

(cherry picked from commit 83ec6e80ff0d56342fc066c2ef649036c0983529)

diff --git a/libstdc++-v3/python/libstdcxx/v6/xmethods.py b/libstdc++-v3/python/libstdcxx/v6/xmethods.py
index 844c8a2105a870fef05fef6cf869795c1b237ff3..8ccf57c4d6b9eb7626a60979e2cc6a4b048e723b 100644 (file)
--- a/libstdc++-v3/python/libstdcxx/v6/xmethods.py
+++ b/libstdc++-v3/python/libstdcxx/v6/xmethods.py
@@ -28,6 +28,8 @@ def get_bool_type():
 def get_std_size_type():
     return gdb.lookup_type('std::size_t')
 
+_versioned_namespace = '__8::'
+
 def is_specialization_of(x, template_name):
     """
     Test whether a type is a specialization of the named class template.