Retry with exponential backoff

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Retry with exponential backoff

Medium·Acceptance —·Asked at Microsoft, Uber, PayPal

Wrap a flaky call so transient failures are retried and permanent ones are not. Return the result, or raise the last exception once the budget is exhausted.

Input schema

solve(call: Callable[[], Any], attempts: int, base_delay: float, retry_on: tuple[type[Exception], ...], sleep: Callable[[float], None]) -> Any

Example

call fails twice with TimeoutError then returns 7 attempts=5, base_delay=0.5, retry_on=(TimeoutError,) sleep called with 0.5 then 1.0; solve returns 7

Constraints

Delay for retry n (1-indexed) is base_delay * 2 ** (n - 1)
Exceptions outside retry_on propagate immediately, with no sleep
The injected sleep is the only delay mechanism — calling time.sleep fails the timing case
attempts counts total calls, not retries

Topics

data modeling