我正在寻找一个宏,如果表达式完成时间超过 X 秒,它将引发异常。
这个问题在这里有更好的答案:执行具有超时功能的函数 https://stackoverflow.com/questions/6694530/executing-a-function-with-a-timeout/6697469#6697469
期货来拯救!
user=> (let [f (future (reduce * (range 1 1001)))]
(.get f 1 java.util.concurrent.TimeUnit/MILLISECONDS))
java.util.concurrent.TimeoutException (NO_SOURCE_FILE:0)
并对其进行宏:
(defmacro time-limited [ms & body]
`(let [f# (future ~@body)]
(.get f# ~ms java.util.concurrent.TimeUnit/MILLISECONDS)))
所以你可以这样做:
user=> (time-limited 1 (reduce * (range 1 1001)))
java.util.concurrent.TimeoutException (NO_SOURCE_FILE:0)
user=> (time-limited 1 (reduce * (range 1 101)))
93326215443944152681699238856266700490715968264381621468592963895217599993229915
608941463976156518286253697920827223758251185210916864000000000000000000000000
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