8

我开始做 99 个 haskell 问题，我在第7 个问题上，我的单​​元测试被炸毁了。

显然，这是由于：http ://www.haskell.org/haskellwiki/Monomorphism_restriction

我只是想确保我理解正确，因为我有点困惑。

情况 1：funca被定义为没有类型 def 或非严格类型 def 然后使用一次，编译器在编译时推断类型没有问题。

情况2：程序中多次使用同一个函数a，编译器不能100%确定类型是什么，除非它为给定的参数重新计算函数。

为了避免计算损失，ghc 向程序员抱怨它需要严格的类型 def ona 才能正常工作。

我认为在我的情况下，assertEqual具有 def 类型

 assertEqual :: (Eq a, Show a) => String -> a -> a -> Assertion

我在定义时遇到错误test3，我将其解释为它有 2 种可能的类型用于返回testcase3(Show 和 Eq) 并且不知道如何继续。

这听起来正确还是我完全关闭了？

问题7.hs：

-- # Problem 7
-- Flatten a nested list structure.

import Test.HUnit

-- Solution

data NestedList a = Elem a | List [NestedList a]

flatten :: NestedList a -> [a]
flatten (Elem x) = [x]
flatten (List x) = concatMap flatten x

-- Tests

testcase1 = flatten (Elem 5)
assertion1 = [5]

testcase2 = flatten (List [Elem 1, List [Elem 2, List [Elem 3, Elem 4], Elem 5]])
assertion2 = [1,2,3,4,5]

-- This explodes
-- testcase3 = flatten (List [])

-- so does this:
-- testcase3' = flatten (List []) :: Eq a => [a]

-- this does not
testcase3'' = flatten (List []) :: Num a => [a]

-- type def based off `:t assertEqual`
assertEmptyList :: (Eq a, Show a) => String -> [a] -> Assertion
assertEmptyList str xs = assertEqual str xs []

test1 = TestCase $ assertEqual "" testcase1 assertion1
test2 = TestCase $ assertEqual "" testcase2 assertion2
test3 = TestCase $ assertEmptyList "" testcase3''

tests = TestList [test1, test2, test3]

-- Main
main = runTestTT tests

第一种情况：testcase3 = flatten (List [])

GHCi, version 7.4.2: http://www.haskell.org/ghc/  :? for help
Loading package ghc-prim ... linking ... done.
Loading package integer-gmp ... linking ... done.
Loading package base ... linking ... done.
[1 of 1] Compiling Main             ( problem7.hs, interpreted )

problem7.hs:29:20:
    Ambiguous type variable `a0' in the constraints:
      (Eq a0)
        arising from a use of `assertEmptyList' at problem7.hs:29:20-34
      (Show a0)
        arising from a use of `assertEmptyList' at problem7.hs:29:20-34
    Probable fix: add a type signature that fixes these type variable(s)
    In the second argument of `($)', namely
      `assertEmptyList "" testcase3'
    In the expression: TestCase $ assertEmptyList "" testcase3
    In an equation for `test3':
        test3 = TestCase $ assertEmptyList "" testcase3
Failed, modules loaded: none.
Prelude>

第二种情况：testcase3 = flatten (List []) :: Eq a => [a]

GHCi, version 7.4.2: http://www.haskell.org/ghc/  :? for help
Loading package ghc-prim ... linking ... done.
Loading package integer-gmp ... linking ... done.
Loading package base ... linking ... done.
[1 of 1] Compiling Main             ( problem7.hs, interpreted )

problem7.hs:22:13:
    Ambiguous type variable `a0' in the constraints:
      (Eq a0)
        arising from an expression type signature at problem7.hs:22:13-44
      (Show a0)
        arising from a use of `assertEmptyList' at problem7.hs:29:20-34
    Possible cause: the monomorphism restriction applied to the following:
      testcase3 :: [a0] (bound at problem7.hs:22:1)
    Probable fix: give these definition(s) an explicit type signature
                  or use -XNoMonomorphismRestriction
    In the expression: flatten (List []) :: Eq a => [a]
    In an equation for `testcase3':
        testcase3 = flatten (List []) :: Eq a => [a]
Failed, modules loaded: none.
4

1 回答 1

4

与其说是单态限制，不如说是默认情况下对歧义类型变量的解析导致编译失败。

-- This explodes
-- testcase3 = flatten (List [])

-- so does this:
-- testcase3' = flatten (List []) :: Eq a => [a]

-- this does not
testcase3'' = flatten (List []) :: Num a => [a]

flatten :: NestedList a -> [a]
flatten (Elem x) = [x]
flatten (List x) = concatMap flatten x

flatten对类型变量没有任何限制a，因此定义本身没有问题testcase3，它将是多态的。

但是当你使用它时test3，

test3 = TestCase $ assertEmptyList "" testcase3 -- ''

你继承了约束

assertEmptyList :: (Eq a, Show a) => String -> [a] -> Assertion

现在编译器必须找出testcase3应该在那里使用哪种类型。没有足够的上下文来确定类型，所以编译器尝试默认解析类型变量。根据默认规则，(Eq a, Show a)无法通过默认解决上下文，因为只有涉及至少一个数字类的上下文才有资格进行默认。因此，由于类型变量不明确，编译失败。

testcase3'testcase3''然而，由于表达式类型签名在定义的右侧施加了由左侧继承的约束，因此属于单态限制。

testcase3'因此无法编译，无论它是否在断言中使用。

testcase3''[Integer]由于表达式类型签名施加了数字约束，因此默认为。因此，当类型为 单态时testcase''，受约束的类型变量默认为Integer。那么在 中使用它的类型就没有问题了test3。

如果您为绑定而不是右侧提供了类型签名，

testcase3' :: Eq a => [a]
testcase3' = flatten (List [])

testcase3'' :: Num a => [a]
testcase3'' = flatten (List [])

这两个值都可以自行编译为多态值，但仍然只能testcase3''在 中使用test3，因为只有这样才引入了允许默认值所需的数字约束。

于 2012-07-29T19:42:44.400 回答