problem with namespaces using eval and exec

[Jens]

Hi,

has anyone an idea why the following code does not work.


s = """
def a(n):
return n*n


def b(t):
return a(t)
"""


ns = {}
exec(s, {}, ns)
eval("b(2)", ns, {})


executing this script raises an exception (NameError: global name 'a'
is not defined) in the last line.


Hope for your help.

 

[Maric Michaud]

Le Mardi 16 Mai 2006 11:05, Jens a écrit :

s = """
def a(n):
  return n*n


def b(t):
  return a(t)
"""


ns = {}
exec(s, {}, ns)
eval("b(2)", ns, {})

you passed an empty globals dictionnary to the eval func (the local one is not
in the scope of b defintion)
try this :

exec s // like exec s in globals(), locals()
eval("b(2)") // like , eval("b(2)", globals(), locals())

or if you to keep this dictionnary as a specific scope in your appplication :

exec s in ns
eval("b(2)", ns)

--
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Maric Michaud
_____________

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Tel: +33 426 880 097

 

[Peter Otten]

has anyone an idea why the following code does not work.

s = """
def a(n):
return n*n

def b(t):
return a(t)
"""

ns = {}
exec(s, {}, ns)

Here you are providing a local namespace into which all toplevel names (a
and b) are inserted.

eval("b(2)", ns, {})

Here you provide a global (the former local) namespace containing an 'a'.
Function b(), however is carrying its global namespace with it and will
look for 'a' in that namespace and not the one you provide. This is similar
to

module x.py
def b(t): return a(t)

module y.py
import x
def a(t): return n * n
x.b(2) # error, will not find the function a() you just defined

Instead you could explicitly update b's global namespace:

global_ns = {}
local_ns = {}
exec s in global_ns, local_ns
global_ns["a"] = local_ns["a"]
print eval("b(2)", local_ns)

or (better, I think) model the situation of an ordinary module where
globals() and locals() are identical:

global_ns = {}
exec s in global_ns
print eval("b(2)", global_ns)

Peter

 

[Jens]

Thanks a lot, that works for me.