Bugs item #1153622, was opened at 2005-02-28 11:48
Message generated for change (Comment added) made by tjreedy
You can respond by visiting:
https://sourceforge.net/tracker/?func=detail&atid=105470&aid=1153622&group_id=5470
Category: Parser/Compiler
Group: Python 2.4
Status: Open
Resolution: None
Priority: 5
Submitted By: Mattias Engdeg�rd (yorick)
Assigned to: Nobody/Anonymous (nobody)
Summary: eval does not bind variables in lambda bodies correctly
Initial Comment:
eval() does not bind variables in lambda expressions
correctly:
>>>def f(g): return eval('lambda x: g(x)')
>>>f(lambda y: y * 2)(17)
Traceback (most recent call last):
File "<stdin>", line 1, in ?
File "<string>", line 1, in <lambda>
NameError: global name 'g' is not defined
The docs say this about eval():
# If both dictionaries are omitted, the expression is
# executed in the environment where eval is called.
and using plain local variables work as expected:
>>>def h(d): return eval('d(10)')
>>>h(lambda y: y * 2)
20
Also, if locals() is presented as the global dict to
eval(), it works:
>>>def f(g): return eval('lambda x: g(x)', locals(),
locals())
>>>f(lambda y: y * 2)(17)
34
but this does not allow the expression to reference
global variables of course.
----------------------------------------------------------------------
Comment By: Terry J. Reedy (tjreedy)
Date: 2005-03-01 12:29
Message:
Logged In: YES
user_id=593130
Whoops. eval('x') == x as code snippets has an exception,
which is the one tripping you up. When the eval is within a
function definition (and lambda expressions are abbreviated
simple function definitions) and 'x' contains a function definition,
then the body of the contained function definition does not have
access, when it is run, to the locals of the containing function
(the lexical scope), whereas it will when x is compiled directly *as
part of the containing function body*. eval('x') removes x from
that part of its context. eval only has the simple two-level
globals/locals environment, which can be anything the caller
passes in, so it compiles x as if it were top-level code. Hence
free variables in contained functions are looked up in the global
passed to eval when the evaled function is called.
This issue has been discussed on the Python newsgroup/mailing
list more than once. If my explanation is not clear, you might be
able to find others in Google c.l.p archives. Do consider that
core functions which have been debugged for over a decade are
unlike to have many bugs left, although the docs are still being
improved.
While Python's scoping is lexical, its free variable binding is late.
Consider
>>> def f():
... x = 0
... def g(): print x
... x = 1
... return g
...
>>> f()()
# What gets printed? 0 or 1?
# From your comments, I suspect you expect 0.
# Irregardless, it is
1
Similarly
>>> f()()
1
>>> d={'x': 0}
>>> h=eval('lambda: x', d, d)
>>> h()
0
>>> d['x'] = 1
>>> h()
# now what gets printed?
1
----------------------------------------------------------------------
Comment By: Mattias Engdeg�rd (yorick)
Date: 2005-03-01 04:11
Message:
Logged In: YES
user_id=432579
>Variables in Python functions are resolved
>when the function is *called*, not when it is defined.
I'm not sure what you mean by that, since Python obeys
lexical scoping, not dynamic.Consider:
def f(x): lambda y: x + y
When the inner lambda expression above is evaluated, x
inside the lambda body is bound to the parameter of the call
of f, even if x+y is not evaluated until that function is
called.
So since
def f(x): return eval('x')
fetches its definition of x from the lexical variable x, why
shouldn't
def f(g): return eval('lambda x: g(x)')
fetch its definition of g from the lexical variable g? A
lambda expression is just a way of delaying evaluation,
*not* delaying how variables are bound --- this is done
immediately.
----------------------------------------------------------------------
Comment By: Terry J. Reedy (tjreedy)
Date: 2005-03-01 00:30
Message:
Logged In: YES
user_id=593130
I am 99.5% sure that this is 'invalid' (a Resolution category) and
should be closed. With the default environment, eval('x') is the
same as unquoted x. Variables in Python functions are resolved
when the function is *called*, not when it is defined. There is no
resolution for g in the default globals. Eval does not change this.
The NameError is exactly correct.
----------------------------------------------------------------------
You can respond by visiting:
https://sourceforge.net/tracker/?func=detail&atid=105470&aid=1153622&group_id=5470
_______________________________________________
Python-bugs-list mailing list
Unsubscribe:
http://mail.python.org/mailman/options/python-bugs-list/archive%40mail-archive.com
