3 ms·
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.
by sdan 7y ago
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.
- BiteCode_dev 7y agoBut no jitter, and no timeout.
- x3ro 7y agoThe 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