From 225b05548027d55aafb11b65f6a4a2bef2f5196f Mon Sep 17 00:00:00 2001 From: Berker Peksag Date: Sun, 19 Aug 2018 13:25:33 +0300 Subject: [PATCH] bpo-22057: Clarify eval() documentation (GH-8812) If a globals dictionary without a '__builtins__' key is passed to eval(), a '__builtins__' key will be inserted to the dictionary: >>> eval("print('__builtins__' in globals())", {}) True (As a result of this behavior, we can use the builtins print() and globals() even if we passed a dictionary without a '__builtins__' key to eval().) --- Doc/library/functions.rst | 6 ++++-- 1 file changed, 4 insertions(+), 2 deletions(-) diff --git a/Doc/library/functions.rst b/Doc/library/functions.rst index e52b0900156..eb41e72ddd3 100644 --- a/Doc/library/functions.rst +++ b/Doc/library/functions.rst @@ -435,8 +435,10 @@ are always available. They are listed here in alphabetical order. The *expression* argument is parsed and evaluated as a Python expression (technically speaking, a condition list) using the *globals* and *locals* dictionaries as global and local namespace. If the *globals* dictionary is - present and lacks '__builtins__', the current globals are copied into *globals* - before *expression* is parsed. This means that *expression* normally has full + present and does not contain a value for the key ``__builtins__``, a + reference to the dictionary of the built-in module :mod:`builtins` is + inserted under that key before *expression* is parsed. + This means that *expression* normally has full access to the standard :mod:`builtins` module and restricted environments are propagated. If the *locals* dictionary is omitted it defaults to the *globals* dictionary. If both dictionaries are omitted, the expression is executed in the