Haskell 使用 UArray 实现数据类型的显示
Haskell implement Show for data type with UArray
我正在尝试为我的数据类型实现 Show 类型类
data Heap a = Heap {invariant :: a -> a -> Ordering
,arr :: UArray Int a}
在此数据上使用显示时,我只希望它显示底层数组。我试过像这样实现表演:
instance Show a => Show (Heap a) where
show (Heap i a) = show a
但这给了我以下错误:
Heap.hs:12:21: error:
• Could not deduce (IArray UArray a) arising from a use of ‘show’
from the context: Show a
bound by the instance declaration at Heap.hs:11:10-32
• In the expression: show a
In an equation for ‘show’: show (Heap i a) = show a
In the instance declaration for ‘Show (Heap a)’
|
12 | show (Heap i a) = show a
| ^^^^^^
我认为问题与参数化类型有关,就像我使用 Int
而不是 a
一样,如下代码工作正常:
data Heap = Heap {invariant :: Int -> Int -> Bool
,arr :: UArray Int Int}
instance Show Heap where
show (Heap i a) = show a
在 GHCI 中
λ > Heap (>) (array (1,10) [])
array (1,10) [(1,0),(2,0),(3,0),(4,0),(5,0),(6,0),(7,0),(8,0),(9,0),(10,0)]
我在使用参数化类型时到底做错了什么?
只需将给定的约束添加到您的实例上下文中。像这样:
instance (IArray UArray a, Show a) => Show (Heap a) where
show (Heap i a) = show a
您可能需要一些语言扩展;编译器会告诉你哪些。
我正在尝试为我的数据类型实现 Show 类型类
data Heap a = Heap {invariant :: a -> a -> Ordering
,arr :: UArray Int a}
在此数据上使用显示时,我只希望它显示底层数组。我试过像这样实现表演:
instance Show a => Show (Heap a) where
show (Heap i a) = show a
但这给了我以下错误:
Heap.hs:12:21: error:
• Could not deduce (IArray UArray a) arising from a use of ‘show’
from the context: Show a
bound by the instance declaration at Heap.hs:11:10-32
• In the expression: show a
In an equation for ‘show’: show (Heap i a) = show a
In the instance declaration for ‘Show (Heap a)’
|
12 | show (Heap i a) = show a
| ^^^^^^
我认为问题与参数化类型有关,就像我使用 Int
而不是 a
一样,如下代码工作正常:
data Heap = Heap {invariant :: Int -> Int -> Bool
,arr :: UArray Int Int}
instance Show Heap where
show (Heap i a) = show a
在 GHCI 中
λ > Heap (>) (array (1,10) [])
array (1,10) [(1,0),(2,0),(3,0),(4,0),(5,0),(6,0),(7,0),(8,0),(9,0),(10,0)]
我在使用参数化类型时到底做错了什么?
只需将给定的约束添加到您的实例上下文中。像这样:
instance (IArray UArray a, Show a) => Show (Heap a) where
show (Heap i a) = show a
您可能需要一些语言扩展;编译器会告诉你哪些。