计划出纯粹的功能 [英] Timing out pure functions

问题描述

我怎样才能杀死耗时过长的纯粹计算？我试过

How can I "kill" a pure calculation which is taking too long? I tried

import System.Timeout

fact 0 = 1
fact n = n * (fact $ n - 1)

main = do maybeNum <- timeout (10 ^ 7) $ (return . fact) 99999999
          print maybeNum

但是，这不起作用。将（return。fact）99999999 替换为一个真正的IO函数，如 getLine ，并按预期工作。 / p>

However, this doesn't work. Replace the (return . fact) 99999999 with a "real" IO function like getLine and this works as expected.

推荐答案

The point is that

return (fact 999999999)

立即返回并且不触发超时。它会返回一个将在稍后评估的thunk。

immediately returns and doesn't trigger the timeout. It returns a thunk that will be evaluated later.

如果您强制评估返回值，那么

If you force evaluation of the return value,

main = do maybeNum <- timeout (10 ^ 7) $ return $! fact 99999999
          print maybeNum

它应该触发超时（如果您提供足够大的堆栈以便在堆栈溢出之前发生超时）。

it should trigger the timeout (if you provide a stack large enough so that the timeout happens before the stack overflow).

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