GHC 7.7中引入的自由覆盖条件打破了GHC 7.6中有效的代码

这个想法

我正在写一个DSL,它编译成Haskell.

该语言的用户可以定义自己的不可变数据结构和相关功能.通过关联函数,我指的是属于数据结构的函数.例如,用户可以写(以"pythonic"伪代码):

data Vector a:
  x,y,z :: a
  def method1(self, x):
      return x

(这相当于下面的代码,但也显示了相关函数beheva类似于具有开放世界假设的类型类):

data Vector a:
  x,y,z :: a
def Vector.method1(self, x):
  return x

在这个例子中,method1是一个与Vector数据类型相关的函数,可以像v.testid(5)(数据类型的v实例Vector)一样使用.

我正在将这样的代码翻译成Haskell代码,但我遇到了一个问题,我试图解决很长一段时间.

问题

我试图将代码从GHC 7.6移到GHC 7.7(预发布7.8)(较新的版本可以从源代码编译).该代码在GHC 7.6下完美运行,但不在GHC 7.7下.我想问你如何修复它以使其在新版本的编译器中工作？

示例代码

让我们看一下生成的简化版本(由我的编译器)Haskell代码:

{-# LANGUAGE FlexibleInstances #-}
{-# LANGUAGE ScopedTypeVariables #-}
{-# LANGUAGE TypeFamilies #-}
{-# LANGUAGE UndecidableInstances #-}
{-# LANGUAGE FunctionalDependencies #-}

import Data.Tuple.OneTuple

------------------------------
-- data types
------------------------------
data Vector a = Vector {x :: a, y :: a, z :: a} deriving (Show)
-- the Vector_testid is used as wrapper over a function "testid". 
newtype Vector_testid a = Vector_testid a

------------------------------
-- sample function, which is associated to data type Vector
------------------------------
testid (v :: Vector a) x = x

------------------------------
-- problematic function (described later)
------------------------------
testx x = call (method1 x) $ OneTuple "test"

------------------------------
-- type classes
------------------------------
-- type class used to access "method1" associated function
class Method1 cls m func | cls -> m, cls -> func where 
    method1 :: cls -> m func

-- simplified version of type class used to "evaluate" functions based on 
-- their input. For example: passing empty tuple as first argument of `call` 
-- indicates evaluating function with default arguments (in this example 
-- the mechanism of getting default arguments is not available)
class Call a b where
    call :: a -> b

------------------------------
-- type classes instances
------------------------------
instance (out ~ (t1->t1)) => Method1 (Vector a) Vector_testid out where
  method1 = (Vector_testid . testid)

instance (base ~ (OneTuple t1 -> t2)) => Call (Vector_testid base) (OneTuple t1 -> t2) where
    call (Vector_testid val) = val

------------------------------
-- example usage
------------------------------
main = do
    let v = Vector (1::Int) (2::Int) (3::Int)
    -- following lines equals to a pseudocode of ` v.method1 "test" `
    -- OneTuple is used to indicate, that we are passing single element.
    -- In case of more or less elements, ordinary tuples would be used.
    print $ call (method1 v) $ OneTuple "test"
    print $ testx v

代码编译并与GHC 7.6一起正常工作.当我尝试使用GHC 7.7编译它时,我收到以下错误:

debug.hs:61:10:
    Illegal instance declaration for
      ?Method1 (Vector a) Vector_testid out’
      The liberal coverage condition fails in class ?Method1’
        for functional dependency: ?cls -> func’
      Reason: lhs type ?Vector a’ does not determine rhs type ?out’
    In the instance declaration for
      ?Method1 (Vector a) Vector_testid out’

该错误是由检查功能依赖性可以做什么的新规则引起的,即liberal coverage condition(据我所知,coverage condition通过使用可以放宽-XUndecidableInstances)

一些人试图解决问题

我试图通过将定义更改Method1为:来克服此问题:

class Method1 cls m func | cls -> m where 
    method1 :: cls -> m func

这解决了功能依赖的问题,但后面的行:

testx x = call (method1 x) $ OneTuple "test"

不再允许,导致编译错误(在7.6和7.7版本中):

Could not deduce (Method1 cls m func0)
  arising from the ambiguity check for ?testx’
from the context (Method1 cls m func,
                  Call (m func) (OneTuple [Char] -> s))
  bound by the inferred type for ?testx’:
             (Method1 cls m func, Call (m func) (OneTuple [Char] -> s)) =>
             cls -> s
  at debug.hs:50:1-44
The type variable ?func0’ is ambiguous
When checking that ?testx’
  has the inferred type ?forall cls (m :: * -> *) func s.
                         (Method1 cls m func, Call (m func) (OneTuple [Char] -> s)) =>
                         cls -> s’
Probable cause: the inferred type is ambiguous

编辑:

使用类型族也是不可能解决这个问题的(据我所知).如果我们Method1用以下代码(或simmilar)替换类型类和实例:

class Method1 cls m | cls -> m where 
    type Func cls
    method1 :: cls -> m (Func cls)

instance Method1 (Vector a) Vector_testid where
    type Func (Vector a) = (t1->t1)
    method1 = (Vector_testid . testid)

我们会得到明显的错误Not in scope: type variable ?t1’,因为类型族不允许使用类型,这种类型不会出现在类型表达式的LHS上.

最后一个问题

如何让这个想法在GHC 7.7下运作？我知道新的liberal coverage condition允许GHC开发人员在类型检查方面取得了一些进展,但它应该能够以某种方式在GHC 7.6中使用从不编译器版本的端口构思.

(不强迫我的DSL用户引入任何其他类型 - 到目前为止的一切,比如类型类实例,我正在使用Template Haskell进行创建)