I looked through the myriad 'Python exec' threads on SO, but couldn't find one that answered my issue. Terribly sorry if this has been asked before. Here's my problem:
# Python 2.6: prints 'it is working'
# Python 3.1.2: "NameError: global name 'a_func' is not defined"
class Testing(object):
def __init__(self):
exec("""def a_func():
print('it is working')""")
a_func()
Testing()
# Python 2.6: prints 'it is working'
# Python 3.1.2: prints 'it is working'
class Testing(object):
def __init__(self):
def a_func():
print('it is working')
a_func()
Testing()
As the standard function definition works in both Python versions, I'm assuming the problem must be a change to the way exec works. I read the API docs for 2.6 and 3 for exec
and also read the "What's New In Python 3.0" page and couldn't see any reason why the code would break.
It's all about flexibility. But the real key to the print function is somewhat subtle and it all has to do with flexibility, both for the users and the Python development team. For users, making print a function lets you use print as an expression, unlike the print statement which can only be used as a statement.
The fundamental difference between Python and Python 3 is that Python 3 is not compatible with the older versions of Python. One significant change is that the code that works in Python 2.7 will not work in Python 3, they need to be written in different ways.
Python 3 is more in-demand and includes a typing system. Python 2 is outdated and uses an older syntax for the print function. While Python 2 is still in use for configuration management in DevOps, Python 3 is the current standard. Python (the code, not the snake) is a popular coding language to learn for beginners.
If you want to determine whether Python2 or Python3 is running, you can check the major version with this sys. version_info. major . 2 means Python2, and 3 means Python3.
You can see the generated bytecode for each Python version with:
>>> from dis import dis
And, for each interpreter:
#Python 3.2
>>> dis(Testing.__init__)
...
5 10 LOAD_GLOBAL 1 (a_func)
...
#Python 2.7
>>> dis(Testing.__init__)
...
5 8 LOAD_NAME 0 (a_func)
...
As you can see, Python 3.2 searches for a global value (LOAD_GLOBAL) named a_func
and 2.7 first searches the local scope (LOAD_NAME) before searching the global one.
If you do print(locals())
after the exec
, you'll see that a_func
is created inside the __init__
function.
I don't really know why it's done that way, but seems to be a change on how symbol tables are processed.
BTW, if want to create a a_func = None
on top of your __init__
method to make the interpreter know it's a local variable, it'll not work since the bytecode now will be LOAD_FAST
and that don't make a search but directly gets the value from a list.
The only solution I see is to add globals()
as second argument to exec
, so that will create a_func
as a global function an may be accessed by the LOAD_GLOBAL
opcode.
Edit
If you remove the exec
statement, Python2.7 change the bytecode from LOAD_NAME
to LOAD_GLOBAL
. So, using exec
, your code will always be slower on Python2.x because it has to search the local scope for changes.
As Python3's exec
is not a keyword, the interpreter can't be sure if it's really executing new code or doing something else... So the bytecode don't change.
E.g.
>>> exec = len
>>> exec([1,2,3])
3
tl;dr
exec('...', globals())
may solve the problem if you don't care the result being added to global namespace
Completing the answer above, just in case. If the exec
is in some function, I would recommend using the three-argument version as follows:
def f():
d = {}
exec("def myfunc(): ...", globals(), d)
d["myfunc"]()
This is the cleanest solution, as it doesn't modify any namespace under your feet. Instead, myfunc
is stored in the explicit dictionary d
.
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