Don't understand syntax error: unqualified exec is not allowed ..

[Stef Mientki]

Terry Reedy wrote:
> Stef Mientki wrote:
>
>>>>    import sys
>>>>    p_locals  = sys._getframe(1).f_locals
>>>
>>> Which locals does this get you? __init__'s? (locals()?)
>> AFAIK locals() == sys._getframe(0).f_locals
>
> Then it is not clear to me which locals your are getting.
The locals of the code block where the instance of this class is created.
>
>
>>>>    p_globals = sys._getframe(1).f_globals
>>>
>>> Isn't this just the same as globals()?
>> AFAIK, again one level up
>>>
>>>>    try :
>>>>      exec ( code, p_globals, p_locals )
>>>
>>> This is 3.0 exec function syntax.
>> weird, I use it in 2.5 and if I remember well it already worked in 2.4.
>> but exchanging the function with the statement yields exactly the 
>> same results
>
> 2.5 manual says
> "6.14 The exec statement
>
> exec_stmt ::= "exec" or_expr ["in" expression ["," expression]]
>
> This statement supports dynamic execution of Python code. The first 
> expression should evaluate to either a string, an open file object, or 
> a code object."
>
> Your expression evaluates to a tuple (triple), which is not documented.
No I think it's indeed handled as function call.
I now that, because the globals (and sometimes ??? the locals) of the 
codeblock where the class is initiated is changed.
> From the error message, I wondoer if p_globals, p_locals is ignored. Try
> exec code in p_globals, p_locals
Yes I've already done that,
and this yields exactly the same results.
So I guess that the exec-function is already there and doesn't differ 
from the exec-statement.
And indeed nothing to find about the exec-function in the 2.5 
documentation :-(

But on the other hand,
I'm a very practical guy,
and my code seems to work quit correct now,
so I'm satisfied for the moment ;-)

thanks again,
Stef
>
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