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I think I’ll try this, but for devs:

a simple while True + try/catch + time.sleep(n*2) will do exponential backoff pretty easily.




The post clearly states the problems they were trying to solve, none of which are taken into account here (consistency, ease of parameterization,..) It seems like people are reading too many "X in Y lines of code" posts :D


But no jitter, and no timeout.




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