[speedster217]

I just tested it in Python3:

    def foo():
        exec("a=1")
        return a

    print(foo())

Fails with a NameError:

    Traceback (most recent call last):
      File "test.py", line 5, in <module>
        print(foo())
      File "test.py", line 3, in foo
        return a
    NameError: name 'a' is not defined

[joncatanio]

I get the same error with your example, but this works fine (Python 3.6.4):

    exec("a = 1")
    print(a)

This will print "1".

[yorwba]

That is because the exec runs in the global scope. When Python sees a variable that is not assigned to in the local scope, it is assumed to be a global variable, so when exec creates a new local variable, the load still fails because it looks into the globals dictionary.

But you can do this:

  def foo():
    exec("a = 1")
    return locals()['a']