Question

我有以下数据

data A = C1 String | A :@: A
         deriving(Show)  

app inp = case inp of
     a1 :@: a2 -> (C1 "a") :@: (C1 "b")
     _         -> C1 "c"

为什么案例会返回输入而不是(C1 "a") :@: (C1 "b")？

*Test> app (C1 "c") :@: (C1 "d")
C1 "c" :@: C1 "d"

如果我将A :@: A更改为C2 A A

，则可以正常工作

Answer 1

函数应用程序的优先级高于:@:（或任何其他中缀运算符），因此app (C1 "c") :@: (C1 "d")与(app (C1 "c")) :@: (C1 "d")相同，而不是app ((C1 "c") :@: (C1 "d"))。后者做你期望的：

*Main> app ((C1 "c") :@@: (C1 "d"))
C1 "a" :@@: C1 "b"

更为惯用的写作方式是app $ (C1 "c") :@: (C1 "d")。