Don't understand syntax error: unqualified exec is not allowed ..
Terry Reedy
tjreedy at udel.edu
Mon Oct 20 18:13:31 EDT 2008
Stef Mientki wrote:
> hello,
>
> I've syntax error which I totally don't understand:
>
> ########## mainfile :
> import test_upframe2
>
> if __name__ == '__main__':
> var1 = 33
> code = 'print var1 + 3 \n'
> test_upframe2.Do_Something_In_Parent_NameSpace ( code )
>
> ########### file = test_upframe2 :
> class Do_Something_In_Parent_NameSpace ( object ) :
> def __init__ ( self, code ) :
> def do_more ( ) :
> nonvar = [3,4]
> while len ( nonvar ) > 0 : # <<<===
> nonvar.pop() # <<<===
Indendation is screwed. Is the above all do_more body?
> import sys
> p_locals = sys._getframe(1).f_locals
Which locals does this get you? __init__'s? (locals()?)
> p_globals = sys._getframe(1).f_globals
Isn't this just the same as globals()?
> try :
> exec ( code, p_globals, p_locals )
This is 3.0 exec function syntax. Postings should specify which
interpreter you are running, especially when mucking with
internals.
> except :
> print 'ERROR'
>
>
> gives me:
> SyntaxError: unqualified exec is not allowed in function '__init__' it
> contains a nested function with free variables (gui_support.py, line
> 408)
> "unqualified exec" : I thought that meant there is some ambiguity in the
> namespace, but I explictly definied the namespace
>
> The function "do_more" is never called, so what does it matter what's
> in there ?
>
> If I remove the while-loop (which of course I can't) the syntax error
> disappears.
>
> I can place the function either as a class method or as a normal
> function outside the class,
> which both works well.
> But I want the method / function not to be hidden.
Since you are hiding it, I presume you mean to be, not not to be.
>
> Why does the above syntax error appear ?
> Are there other ways to hide the function ?
Either use module level __all__ or name the function _do_more and anyone
will know it is private to the module.
More information about the Python-list
mailing list