Haskell:用两个 float 参数组合函数失败

Question

我正在尝试编写 (Floating a) => a -> a -> a 类型的函数具有 (Floating a) => a -> a 类型的函数获得 (Floating a) => a -> a -> a 类型的函数.我有以下代码:

test1 :: (Floating a) => a -> a -> a
test1 x y = x

test2 :: (Floating a) => a -> a
test2 x = x

testBoth :: (Floating a) => a -> a -> a
testBoth = test2 . test1
--testBoth x y = test2 (test1 x y)

但是，当我在 GHCI 中编译它时，出现以下错误:

/path/test.hs:8:11:
    Could not deduce (Floating (a -> a)) from the context (Floating a)
      arising from a use of `test2'
                   at /path/test.hs:8:11-15
    Possible fix:
      add (Floating (a -> a)) to the context of
        the type signature for `testBoth'
      or add an instance declaration for (Floating (a -> a))
    In the first argument of `(.)', namely `test2'
    In the expression: test2 . test1
    In the definition of `testBoth': testBoth = test2 . test1
Failed, modules loaded: none.

注意 testBoth 的注释掉的版本编译。奇怪的是，如果我删除 (Floating a)所有类型签名的约束或如果我更改 test1就拿 x而不是 x和 y , testBoth编译。

我搜索了 StackOverflow、Haskell wikis、Google 等，但没有发现任何与此特定情况相关的函数组合限制。有谁知道为什么会这样？

Answer 1

   \x y -> test2 (test1 x y)
== \x y -> test2 ((test1 x) y)
== \x y -> (test2 . (test1 x)) y
== \x -> test2 . (test1 x)
== \x -> (test2 .) (test1 x)
== \x -> ((test2 .) . test1) x
== (test2 .) . test1

这两件事彼此不一样。

   test2 . test1
== \x -> (test2 . test1) x
== \x -> test2 (test1 x)
== \x y -> (test2 (test1 x)) y
== \x y -> test2 (test1 x) y