Are decorators really that different from metaclasses...
Paul Morrow
pm_mon at yahoo.com
Tue Aug 31 08:47:38 EDT 2004
The big impact of this proposal is that __xxx__ variables defined
immediately following the docstring (if present) would cease to become
local variables. This of course has the possiblity of breaking existing
code. But I don't believe that developers routinely use such names for
local variables, so I don't believe that there will actually be much
broken code should this change be implemented.
Jeff Epler wrote:
> In your proposed model, what does the following code do?
> def f():
> __var__ = 3
> return __var__
> f.__var__ += 1
> print f()
>
Only assignments to __xxx__ variables at the very start of a function
def would have this special semantics. So your return statement would
be referencing a name (__var__) that doesn't exist in the function's
local variable namespace.
Easier to accept if we use a more likely variable name for the magic
variable.
def f():
__version__ = 3
return __version__
f.__version__ += 1
print f()
> What about any of the following:
> def g():
> if True:
> __var__ = 4
> print g.__var__
>
__var__ = 4 # creates a local variable because
# the assignment isn't at top
# of function def. But probably
# should generate a warning for
# giving a magic name to a local
# variable.
> x = 3
> def h(x):
> __var__ = x*x
> return x*x
> print h(2), h.__var__
>
In __var__ = x*x, x is not in the function's namespace (it's a local
variable), so x is undefined. Easier to visualize if we make it
__version__ = x*x
Same basic idea for your other examples.
>
> def p():
> import os as __os__
>
__os__ would be a local variable (since this is not a simple assignment,
i.e. using an equal sign). __os__ is not a good name for a local
variable, so interpretor should probably generate a warning.
> def q(): # BUG x doesn't get the proper metaclass in 2.3!
> class __metaclass__(type): pass
> class x: pass
> # assert x's metaclass is __metaclass__
>
This would do whatever it does now.
Paul
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