浅谈Frege下的教会数字程序

Question

此程序在GHC下正确编译和运行：

type Church a = (a -> a) -> a -> a

ch :: Int -> Church a
ch 0 _ = id
ch n f = f . ch (n-1) f

unch :: Church Int -> Int
unch n = n (+1) 0

suc :: Church a -> Church a
suc n f = f . n f

pre :: Church ((a -> a) -> a) -> Church a
pre n f a = n s z id
    where s g h = h (g f)
          z     = const a

main :: IO ()
main = do let seven = ch 7
              eight = suc seven
              six   = pre seven
          print (unch eight)
          print (unch six)

但是，在使用Frege编译时，我得到了以下错误：

E /home/xgp/work/flab/src/main/frege/flab/fold.fr:23: type error in expression seven
    type is : Int
    expected: (t1→t1)→t1
E /home/xgp/work/flab/src/main/frege/flab/fold.fr:23: type error in expression seven
    type is : (t1→t1)→t1
    expected: Int
E /home/xgp/work/flab/src/main/frege/flab/fold.fr:23: type error in expression seven
    type is : (t1→t1)→t1
    expected: Int
E /home/xgp/work/flab/src/main/frege/flab/fold.fr:23: type error in
    expression seven
    type is apparently Int
    used as function

为什么？是否可以修改程序以通过Frege下的编译？

Answer 1

这是一种罕见的情况，在这种情况下，泛化类型的let绑定变量确实会产生影响。

问题是，弗雷格在这方面的表现就像GHC一样，关于语用-XMonoLocalBinds的详细信息，请参阅此处：https://github.com/Frege/frege/wiki/GHC-Language-Options-vs.-Frege#Let-Generalization和guide/other-type-extensions.html#typing-binds (还有一个链接到SPJ的一篇论文，其中解释了理由)。

简而言之，这意味着所有未注释的让绑定验证都将具有一个单一类型，并且不能在不同的类型中使用。要恢复多态性，需要显式类型签名。

要使您的程序编译，只需用

seven :: Church a

关于print/println:前者不刷新输出。所以你在REPL里有：

frege> print 'a'
IO ()
frege> print 'b'
IO ()
frege> println "dammit!"
abdammit!
IO ()