评论者:pbrown
我还遇到了另一种额外嵌套的情况,当重复一个或多个序列并在开始或结束处有一个可选元素时,该元素的谓词匹配另一端元素
user=> (s/conform (s/+ (s/cat :k any? :v (s/? any?))) [:a 1 :b 2]) [{:k :a, :v 1} [{:k :b, :v 2}]]
而我期望的是
`
[{:k :a, :v 1} {:k :b, :v 2}] `
以下给出了期望的结果
user=> (s/conform (s/+ (s/cat :k any? :v (s/? any?))) [:a 1 :b]) [{:k :a, :v 1} {:k :b}] user=> (s/conform (s/+ (s/cat :k keyword? :v (s/? int?))) [:a 1 :b 2]) [{:k :a, :v 1} {:k :b, :v 2}] user=> (s/conform (s/* (s/cat :k any? :v (s/? any?))) [:a 1 :b 2]) [{:k :a, :v 1} {:k :b, :v 2}]