能给我解释一下单态限制吗？

Question

我开始做99个haskell问题，我在problem 7上，我的单元测试失败了。

显然，这是由于以下原因：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: func a被定义为没有类型定义或非严格类型定义，然后使用一次，编译器在编译时推断类型没有问题。

情况2:相同的函数a在程序中被多次使用，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向程序员抱怨说，它需要在a上定义严格的类型才能正常工作。

我认为在我的情况下，assertEqual的类型定义为

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

当定义test3时，我得到了一个错误，我解释说它有两种可能的testcase3返回类型(Show和Eq)，并且不知道如何继续。

这听起来对吗?还是我完全错了？

problem7.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.

Answer 1

与其说是单态限制，不如说是defaulting对不明确类型变量的解析导致了编译失败。

-- 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没有对类型变量a施加任何约束，因此testcase3的定义本身没有问题，它是多态的。

但是当你在test3中使用它时，

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

继承了的约束

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

现在编译器必须找出在那里应该使用哪种类型的testcase3。没有足够的上下文来确定类型，因此编译器尝试通过缺省来解析类型变量。根据defaulting rules，上下文(Eq a, Show a)不能通过缺省来解析，因为只有包含至少一个数字类的上下文才有资格缺省。因此，由于类型变量不明确，编译失败。

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testcase3'和testcase3''受到单态限制。

无论是否在断言中使用，testcase3'都会因此而无法编译。

由于表达式类型签名施加了数字约束，因此testcase3''默认设置为[Integer]。因此，当类型是testcase''的单形化时，约束类型变量缺省为Integer。那么就不存在在test3中使用它的类型的问题了。

如果将类型签名提供给绑定，而不是提供给右侧，

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

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

这两个值都会自行编译成多态值，但在test3中仍然只有testcase3''可用，因为只有这样才会引入所需的数字约束来允许缺省。