On 9/4/2012 4:28 PM, channel727...@gmail.com wrote:
The Python C API function PyEval_EvalCode let's you execute compiled
Python code. I want to execute a block of Python code as if it were
executing within the scope of a function, so that it has its own
dictionary of local variables which don't affect the global state.
You cannot really do this. The local namespace of functions is not a dict.
This seems easy enough to do, since PyEval_EvalCode lets you provide
a Global and Local dictionary:
PyObject* PyEval_EvalCode(PyCodeObject *co, PyObject *globals,
PyObject *locals)
When you do this, you are executing code as if in a class statement, not
as if in a function.
The problem I run into has to do with how Python looks up variable
names. Consider the following code, that I execute with
PyEval_EvalCode:
myvar = 300
> def func(): return myvar
func()
if you wrapped this in def outer(), myvar would be a nonlocal variable,
neither local nor global. exec does not allow for that.
This simple code actually raises an error, because Python is unable
to find the variable myvar from within func.
This is the same as if you indented the above under class xxx:
...
It's only with nested functions, that Python seems to copy
outer-scope locals into inner-scope locals.
I was surprised to find that nonlocals do appear in locals(), but I
would not exactly say that they are copied into the actual local namespace.
def outer():
x = 1
def inner(z=3):
y = 2
print(x)
print(locals())
inner()
print(inner.__code__.co_varnames, inner.__code__.co_freevars)
outer()
#
1
{'x': 1, 'y': 2, 'z': 3}
('z', 'y') ('x',)
Varnames are the local names, freevars are the nonlocal names.
In any case, this complexity requires the presence of an outer function
when the code is compiled. There is no way to simulate it after the fact
by fiddling with just locals and globals when the compiled code is
exec'ed. Class statement, on the other hand, simply introduce a local
namespace separate from the module namespace.
--
Terry Jan Reedy
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