"gangesmaster" <[EMAIL PROTECTED]> wrote: > what problem does the cell object solve?
The closure represents the variable, not the object. So if x is rebound to
a different object your inner function g() will now access the new object.
If the object itself was passed to MAKE_CLOSURE then g would only ever see
the value of x from the instant when g was defined.
>>> def f(x):
def g():
print "x is", x
g()
x += 1
g()
>>> f(1)
x is 1
x is 2
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