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我在想如何強制一個懶惰的函數序列進行評估。執行一個懶惰的函數序列
舉例來說,如果我有一個返回整數1功能:
test.core=> (fn [] 1)
#<core$eval2480$fn__2481 [email protected]>
test.core=> ((fn [] 1))
1
我構建這些功能的懶惰序列:
test.core=> (repeat 5 (fn [] 1))
(#<core$eval2488$fn__2489 [email protected]> ...)
test.core=> (class (repeat 5 '(fn [] 1)))
clojure.lang.LazySeq
我怎麼竟在執行功能序列?
test.core=> (take 1 (repeat 5 (fn [] 1)))
(#<core$eval2492$fn__2493 [email protected]>)
test.core=> (take 1 (repeat 5 '(fn [] 1)))
((fn [] 1))
test.core=> ((take 1 (repeat 5 '(fn [] 1))))
ClassCastException clojure.lang.LazySeq cannot be cast to clojure.lang.IFn
我已經通過閱讀How to convert lazy sequence to non-lazy in Clojure,這表明的doall ......但我不知道在哪裏,結果去?我期待[1 1 1 1 1]或類似的東西。
test.core=> (doall (repeat 5 (fn [] 1)))
(#<core$eval2500$fn__2501 [email protected]>...)
test.core=> (realized? (doall (repeat 5 (fn [] 1))))
true
而且值得注意的是,序列分塊會導致它們以塊爲單位進行評估。12 –
僅適用於產生分塊序列的函數(「重複」不)。分塊問題偶爾會發生,但在OP計算出如何調用函數列表之後很長時間內可以忽略它。 – amalloy