Michael Spencer wrote:
def fun(dict): # set dict as local namespace # locals() = dict? z = x + y
As you no doubt have discovered from the docs and this group, that isn't doable with CPython.
Not entirely impossible:
Py> def f(d): ... exec "locals().update(d)" ... return x + y ... Py> f(dict(x=1, y=2)) 3
Due to the way 'exec' is implemented, modifications to locals() inside an exec statement actually take effect (basically, they're freeloading on the code which allows 'exec "x = 1"' to work properly).
This is an evil, evil hack and almost certainly not what anyone should be doing. Also, variables created this way will be slower than normal variables due to the way the associated code works.
Cheers, Nick.
Oooh - evil indeed, but thanks for the pointer.
I debated including a link to one of the 'writable locals' threads, when I settled on not 'doable', but gambled on being probably useful rather than certainly accurate. Just goes to show you can't get away with anything in this NG ;-)
Cheers
Michael
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