From: R David Murray <rdmurray@bitdance.com>
Date: Tue, 21 May 2013 15:44:41 +0000 (-0400)
Subject: #17973: fix technical inaccuracy in faq entry (it now passes doctest).
X-Git-Tag: v3.4.0a1~658^2
X-Git-Url: http://git.ipfire.org/gitweb.cgi?a=commitdiff_plain;h=95ae99205e3a9da45b958415af92432ca990914e;p=thirdparty%2FPython%2Fcpython.git

#17973: fix technical inaccuracy in faq entry (it now passes doctest).
---

diff --git a/Doc/faq/programming.rst b/Doc/faq/programming.rst
index 7713450cfe97..6a720d10319b 100644
--- a/Doc/faq/programming.rst
+++ b/Doc/faq/programming.rst
@@ -1131,7 +1131,7 @@ a tuple points to.
 Under the covers, what this augmented assignment statement is doing is
 approximately this::
 
-   >>> result = a_tuple[0].__iadd__(1)
+   >>> result = a_tuple[0] + 1
    >>> a_tuple[0] = result
    Traceback (most recent call last):
      ...
@@ -1154,16 +1154,19 @@ that even though there was an error, the append worked::
     >>> a_tuple[0]
     ['foo', 'item']
 
-To see why this happens, you need to know that for lists, ``__iadd__`` is equivalent
-to calling ``extend`` on the list and returning the list.  That's why we say
-that for lists, ``+=`` is a "shorthand" for ``list.extend``::
+To see why this happens, you need to know that (a) if an object implements an
+``__iadd__`` magic method, it gets called when the ``+=`` augmented assignment
+is executed, and its return value is what gets used in the assignment statement;
+and (b) for lists, ``__iadd__`` is equivalent to calling ``extend`` on the list
+and returning the list.  That's why we say that for lists, ``+=`` is a
+"shorthand" for ``list.extend``::
 
     >>> a_list = []
     >>> a_list += [1]
     >>> a_list
     [1]
 
-is equivalent to::
+This is equivalent to::
 
     >>> result = a_list.__iadd__([1])
     >>> a_list = result