Question

请考虑以下功能，即使实施不相关：

vectTranspose : Vect n (Fin 3) -> Vect 3 (List (Fin n))
vectTranspose {n = Z} [] = [[],[],[]]
vectTranspose {n = (S len)} (x :: xs) with (natToFin len (S len))
    | Just l = let
        previous = map (map weaken) (vectTranspose xs)
    in updateAt x (l ::) previous

如果我尝试在REPL中计算vectTranspose，我会按预期得到[[],[],[]]。

但是，如果我在我的代码中添加如下所示的相等断言

emptyTest : vectTranspose [] = [[],[],[]]
emptyTest = Refl

然后我收到编译错误：

When checking right hand side of emptyTest with expected type
        vectTranspose [] = [[], [], []]

Type mismatch between
        [[], [], []] = [[], [], []] (Type of Refl)
and
        vectTranspose [] = [[], [], []] (Expected type)

Specifically:
        Type mismatch between
                [[], [], []]
        and
                vectTranspose []

我错过了什么吗？我应该在断言中以某种方式指定[[],[],[]]的类型吗？

Answer 1

idris抱怨

的类型

Specifically:
    Type mismatch between
            [[], [], []]
    and
            vectTranspose []

表示vectTranspose仍在类型中且尚未解析。如果vectTranspose不是完全的，实际上它不是：

VecTest.vectTranspose is possibly not total due to:
with block in VecTest.vectTranspose, which is not total as there are missing cases

这是因为您没有涵盖所有Maybe个案例。

一个简单的解决方案可能是创建一个小辅助函数：

total
natToFin': (n: Nat) -> Fin (S n)
natToFin' Z = FZ
natToFin' (S k) = FS (natToFin' k)

total
vectTranspose : Vect n (Fin 3) -> Vect 3 (List (Fin n))
vectTranspose {n = Z} [] = [[], [], []]
vectTranspose {n = (S len)} (x :: xs) with (natToFin' len)
    vectTranspose {n = (S len)} (x :: xs) | l = let
           previous = map (map weaken) (vectTranspose xs)
           in updateAt x (l ::) previous

伊德里斯 - 平等断言失败

1 个答案: