Is Python's dict.pop atomic?

Question

It seems reasonable to believe that dict.pop operates atomically, since it raises KeyError if the specified key is missing and no default is provided, like so:

d.pop(k)

However, the documentation does not appear to specifically address that point, at least not in the section specifically documenting dict.pop.

This question occurred to me as I was reviewing an answer of mine which used this pattern:

if k in d: del d[k]

At the time, I was not thinking of the potential condition that a key may be present during the if, but not at the time of del. If dict.pop does indeed provide an atomic alternative, then I should note that in my answer.

Answer 1

For the default type, dict.pop() is a C-function call, which means that it is executed with one bytecode evaluation. This makes that call atomic.

Python threads switch only when the bytecode evaluation loop lets them, so at bytecode boundaries. Some Python C functions do call back into Python code (think __dunder__ special method hooks), but the dict.pop() method does not, at least not for the default dict type.