I am seeing something strange with execfile. I've simplified the code
to:
########## t.py ##########
print "here"
v = None
def f():
global v
v = 6
if __name__ == "__main__":
f()
print "0:", v
execfile("x.py" )
print "0:", v
execfile("y.py" )
print "0:", v
########## x.py and y.py (they are identical) ##########
import testexec
print "1:", v, testexec.v
testexec.v = 7
Runing "python t.py" (with Python 2.4.2), the printout I got is:
here
0: 6
here
1: 6 None
0: 6
1: 6 7
0: 6
So there is apparently 2 different instances of v at run time. Can
someone please explain (1) why this is the case, and (2) assuming this
is correct behavior, how I can avoid this? Thanks. 2 1479
> ########## x.py and y.py (they are identical) ########## import testexec print "1:", v, testexec.v testexec.v = 7
Oops, testexec should be changed to t instead. That is:
########## x.py and y.py (they are identical) ##########
import t
print "1:", v, t.v
t.v = 7
Dennis Lee Bieber wrote: And the problem you are seeing is that the initial "v" in the t.py that you "run", is considered "__main__.v ", NOT "t.v"
Yes, the 2 different copies of v apparently imply that __main__ and t
are 2 different modules. But I had expected __main__ to be an alias of
t. Can you point out the passage in Python doc that explains this? This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics |
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