Issue 8818: urlparse/urlsplit keyword is 'scheme', not 'default_scheme'.

Doc/library/urlparse.rst

@@ -36,7 +36,7 @@ following URL schemes: ``file``, ``ftp``, ``gopher``, ``hdl``, ``http``,
 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 @@ The :mod:`urlparse` module defines the following functions:
       >>> 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 @@ The :mod:`urlparse` module defines the following functions:
    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