How does the eval() function change the dict?
dictionary, eval, python, python-3.x
Solution
The second argument to `eval()` is the globals used for the expression run by `eval()`.
One of things python does when evaluating an expression is ensuring that the python built-ins are available to the evaluated expression, and to do that it adds the `__builtins__` entry to that globals namespace.
So, yes, the `eval()` call did change your dictionary, and that is expected and normal behaviour. It even says so in the documentation for the function:
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 access to the standard `__builtin__` module and restricted environments are propagated.
If you want to avoid this change, use an empty `dict` for the globals, and use `a` as the `locals` namespace instead:
eval('a == "a"', {}, a)
Problem
How does eval() change the dict? This is an example: create a dict -> print -> eval -> print ``` >>> a={'a':'a','b':'b'} >>> print(a) {'a': 'a', 'b': 'b'} >>> eval('a == "a"',a) True >>> print(a) {'a': 'a', '__builtins__': {'bytearray': <class 'bytearray'>, 'IndexError': <class 'IndexError'>, 'all': <built-in function all>, 'help': Type help() for interactive help, or help(object) for help about object., 'vars': <built-in function vars>, 'SyntaxError': <class 'SyntaxError'>, 'UnicodeDecodeError': <class 'UnicodeDecodeError'>, 'memoryview': <class 'memoryview'>, 'isinstance': <built-in function isinstance>, '__build_class__': <built-in function __build_class__>, 'copyright': Copyright (c) 2001-2012 Python Software Foundation. All Rights Reserved. ... ```