Ian Kelly wrote:
Sure, but that's not actually out of sync. The argument of your exec
evaluates to 'print (a)'. You get two different results because
you're actually printing two different variables.
Ah -- thanks, I missed that.
You can get the dict temporarily out of sync:
def f(x, y):
... frob = None
... loc = locals()
... loc[x] = y
... print(loc)
... print(locals())
... print(loc)
...
f('frob', 42)
{'y': 42, 'x': 'frob', 'frob': 42, 'loc': {...}}
{'y': 42, 'x': 'frob', 'frob': None, 'loc': {...}}
{'y': 42, 'x': 'frob', 'frob': None, 'loc': {...}}
In this case, 'frob' is updated to 42 in the dict, but the optimized
local is not updated. Calling locals() again refreshes the dict.
I'm not sure what you mean by temporary:
--> def f(x, y):
... frob = None
... loc = locals()
... loc[x] = y
... print(loc)
... print(locals())
... print(loc)
... print(locals())
...
-->
--> f('frob', 19)
{'y': 19, 'x': 'frob', 'frob': 19}
{'y': 19, 'x': 'frob', 'frob': None, 'loc': {...}}
{'y': 19, 'x': 'frob', 'frob': None, 'loc': {...}}
{'y': 19, 'x': 'frob', 'frob': None, 'loc': {...}}
Seems to be stuck that way.
Here is a better example I was thinking of:
--> def f(x, y):
... locals()[x] = y
... locals()['x'] = 17
... print(locals())
... print(x)
... print(y)
...
--> f('a', 42)
{'y': 42, 'x': 'a', 'a': 42}
a
42
So locals() was updated with 'a', but not with the assignment to 'x'.
And of course, if we tried to 'print(a)' we'd get a NameError.
~Ethan~
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