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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