如何在规范中为两个不同的路径生成相同的值？

您可以使用test.check执行此操作fmap:

(s/def ::a (s/int-in 1 5))
(s/def ::b (s/int-in 1 6))
(s/def ::child (s/keys :req-un [::a ::b]))
(s/def ::parent (s/keys :req-un [::a ::b ::child]))
(gen/sample
  (s/gen ::parent
         {::parent ;; override default gen with fmap'd version
          #(gen/fmap
            (fn [{:keys [a b child] :as p}]
              (assoc p :child (assoc child :a a :b b)))
            (s/gen ::parent))}))
=>
({:a 1, :b 2, :child {:a 1, :b 2}}
 {:a 2, :b 2, :child {:a 2, :b 2}}
 {:a 1, :b 1, :child {:a 1, :b 1}}
 {:a 3, :b 2, :child {:a 3, :b 2}}
 {:a 2, :b 4, :child {:a 2, :b 4}}
 {:a 4, :b 4, :child {:a 4, :b 4}}
 {:a 3, :b 3, :child {:a 3, :b 3}}
 {:a 4, :b 4, :child {:a 4, :b 4}}
 {:a 3, :b 4, :child {:a 3, :b 4}}
 {:a 3, :b 4, :child {:a 3, :b 4}})

fmap接受一个函数f和一个生成器gen,并返回一个适用f于生成的每个值的新生成器gen.在这里,我们传递默认生成器::parent,以及一个获取这些父映射并将相应键复制到映射中的函数:child.

如果您希望此规范强制执行该等式(除了生成),您还需要使用谓词添加s/and到::parent规范中以检查:

(s/def ::parent
  (s/and (s/keys :req-un [::a ::b ::child])
         #(= (select-keys % [:a :b])
             (select-keys (:child %) [:a :b]))))

编辑:这是另一种做同样事情的方法,gen/let它允许更"自然" let的语法:

(gen/sample
  (gen/let [{:keys [a b] :as parent} (s/gen ::parent)
            child (s/gen ::child)]
    (assoc parent :child (assoc child :a a :b b))))