On 5/27/2019 3:18 AM, Greg Ewing wrote:
Chris Angelico wrote:
Except that it does. After calling locals() a second time, the result
of the *first* call will be updated to reflect changes.
Yeow. That's *really* unintuitive. There had better be an extremely
good reason for this behaviour.
I believe that the situation is or can be thought of as this: there is
exactly 1 function locals dict. Initially, it is empty and inaccessible
(unusable) from code. Each locals() call updates the dict to a current
snapshot and returns it.
--
Terry Jan Reedy
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