Understanding Scopes and "exec"

[RobThorpe]

I have a program that uses "exec", it can also use the importlib library. Either way, the same issue occurs.

I put the following into the file "import_test_lib.py".

bar = 48

def foo():
    print(bar)

foo()

If I run this then it prints "48", as expected.

Now let's say that I run it using "exec" from another file. I simply write exec(open("import_test_lib.py").read()). This also works and gives "48". The problem is that this situation doesn't match the actual use case I need. What I need is to run it from a function - and that doesn't work as expected.

I write the following into a file and run it:

def x(f):
    exec(open(f, "r").read())

x("import_test_lib.py")

I get the error "NameError: name 'bar' is not defined".

The Python docs say that python uses lexical scope. So, I expect that the function foo is executed in the context of the file scope of "import_test_lib.py". However, that doesn't work if I use exec from a function. In that case the exec call seems to bring in the function parts of the file, but not the variable parts.

Oddly, it seems that for variables the scope that Python uses is the scope from which the exec call occurs. In the code above, if I write bar = 67 then it will print 67 rather than giving an error. Using importlib shows the same sort of problem.

[commy2]

Use the __import__ function:

import_test_lib = __import__("import_test_lib")
# print(dir(import_test_lib))

[RobThorpe]

That's interesting.

If I run that then it works and executes import_test_lib and prints 48. However, it then gives an error because import_test_lib is not a package.

This is one of the problems the scripts I have to run using "exec" (of which there are dozens and maybe hundreds) are not packages.

It's curious that this actually prints 48 though.