If statement 如何在测试和cond结果中使用结果?
我有以下代码:If statement 如何在测试和cond结果中使用结果?,if-statement,clojure,If Statement,Clojure,我有以下代码: (cond (case-1? (compute-1 x)) (compute-1 x) (case-2? (compute-2 x)) (compute-2 x) (case-3? (compute-3 x)) (compute-3 x)) 我希望避免重复计算compute-1、compute-2和compute-3。一种选择是: (let [result-1 (compute-1 x) result-2 (compute-2 x) resu
(cond
(case-1? (compute-1 x)) (compute-1 x)
(case-2? (compute-2 x)) (compute-2 x)
(case-3? (compute-3 x)) (compute-3 x))
我希望避免重复计算compute-1
、compute-2
和compute-3
。一种选择是:
(let [result-1 (compute-1 x)
result-2 (compute-2 x)
result-3 (compute-3 x)]
(cond
(case-1? result-1) result-1
(case-2? result-2) result-2
(case-3? result-3) result-3))
现在我不再重复计算,而是如果现在(case-1?result-1)
的计算结果为true,result-2
和result-3
是无理由计算的。行为上我想要这样的东西:
(let [result-1 (compute-1 x)]
(if (case-1? result-1)
result-1
(let [result-2 (compute-2 x)]
(if (case-2? result-2)
result-2
(let [result-3 (compute-3 x)]
(if (case-3? result-3)
result-3))))))
(ns foo.core
(:require
[better-cond.core :as b]))
(b/cond
:let [result-1 (compute-1 x)]
(case-1? result-1) result-1
:let [result-2 (compute-2 x)]
(case-2? result-2) result-2
:let [result-3 (compute-3 x)]
(case-3? result-3) result-3)
然而,这段代码显然很快变得难以管理。这个问题有更好的解决方案吗?有一个名为的库,它可以用宏精确地解决这个问题
它是这样使用的:
(let [result-1 (compute-1 x)]
(if (case-1? result-1)
result-1
(let [result-2 (compute-2 x)]
(if (case-2? result-2)
result-2
(let [result-3 (compute-3 x)]
(if (case-3? result-3)
result-3))))))
(ns foo.core
(:require
[better-cond.core :as b]))
(b/cond
:let [result-1 (compute-1 x)]
(case-1? result-1) result-1
:let [result-2 (compute-2 x)]
(case-2? result-2) result-2
:let [result-3 (compute-3 x)]
(case-3? result-3) result-3)
该片段将宏扩展为类似于上一个示例的代码 如果不需要宏或外部dep,可以使用
delay
来避免计算。