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author | Georg Brandl <georg@python.org> | 2009-02-21 19:09:40 (GMT) |
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committer | Georg Brandl <georg@python.org> | 2009-02-21 19:09:40 (GMT) |
commit | a12a86e956f949bd145cef247a95ed742386c70b (patch) | |
tree | f086140b605582498d98b327e3694e79d967b558 /Doc/c-api/intro.rst | |
parent | be9b765c073eefcc109320b651d977ff03090f2f (diff) | |
download | cpython-a12a86e956f949bd145cef247a95ed742386c70b.zip cpython-a12a86e956f949bd145cef247a95ed742386c70b.tar.gz cpython-a12a86e956f949bd145cef247a95ed742386c70b.tar.bz2 |
#5338, #5339: two types in the API manual.
Diffstat (limited to 'Doc/c-api/intro.rst')
-rw-r--r-- | Doc/c-api/intro.rst | 2 |
1 files changed, 1 insertions, 1 deletions
diff --git a/Doc/c-api/intro.rst b/Doc/c-api/intro.rst index 9befaf7..e8a3fc4 100644 --- a/Doc/c-api/intro.rst +++ b/Doc/c-api/intro.rst @@ -187,7 +187,7 @@ caller is said to receive a *new* reference. When no ownership is transferred, the caller is said to *borrow* the reference. Nothing needs to be done for a borrowed reference. -Conversely, when a calling function passes it a reference to an object, there +Conversely, when a calling function passes in a reference to an object, there are two possibilities: the function *steals* a reference to the object, or it does not. *Stealing a reference* means that when you pass a reference to a function, that function assumes that it now owns that reference, and you are not |