Issue 8818: urlparse/urlsplit keyword is 'scheme', not 'default_scheme'.
diff --git a/Doc/library/urlparse.rst b/Doc/library/urlparse.rst
index 135f46c..1f58023 100644
--- a/Doc/library/urlparse.rst
+++ b/Doc/library/urlparse.rst
@@ -36,7 +36,7 @@
The :mod:`urlparse` module defines the following functions:
-.. function:: urlparse(urlstring[, default_scheme[, allow_fragments]])
+.. function:: urlparse(urlstring[, scheme[, allow_fragments]])
Parse a URL into six components, returning a 6-tuple. This corresponds to the
general structure of a URL: ``scheme://netloc/path;parameters?query#fragment``.
@@ -58,7 +58,7 @@
>>> o.geturl()
'http://www.cwi.nl:80/%7Eguido/Python.html'
- If the *default_scheme* argument is specified, it gives the default addressing
+ If the *scheme* argument is specified, it gives the default addressing
scheme, to be used only if the URL does not specify one. The default value for
this argument is the empty string.
@@ -161,7 +161,7 @@
equivalent).
-.. function:: urlsplit(urlstring[, default_scheme[, allow_fragments]])
+.. function:: urlsplit(urlstring[, scheme[, allow_fragments]])
This is similar to :func:`urlparse`, but does not split the params from the URL.
This should generally be used instead of :func:`urlparse` if the more recent URL