我试图编写一个类型为（Floating a）=>的函数。 a - > a - >一个带有类型的函数的（Floating a）=> a - > a 来获得类型为（Floating a）=>的函数。 a - > a - >一个。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 - > testBoth = test2。 test1 --testBoth xy = test2（test1 xy）然而，当我编译它时在GHCI中，我得到以下错误：
/path/test.hs:8:11：无法从/path/test.hs:8:11-15 $ b $处使用`test2'引起的上下文（Floating a）中的浮点数（a-> a） b可能的修正：将 add（浮动（a - > a））添加到的上下文中`testBoth'的类型签名或者添加一个实例声明（Floating（a - > ; a））在`（。）'的第一个参数中，即`test2'在表达式：test2中。 test1 在`testBoth'的定义中：testBoth = test2。 test1 失败，模块加载：无。请注意，注释版本 testBoth 编译。奇怪的是，如果我从所有类型签名中删除（Floating a）约束，或者如果将 test1 更改为只需 x 而不是 x 和 y ， testBoth 编译。
我搜索了StackOverflow，Haskell wiki，Google等，功能组成与这种特定情况有关。有人知道为什么会这样吗？
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这两样东西并不相同。
test2。 test1 == \x - > （test2。test1）x == \x - > test2（test1 x） == \x y - > （test2（test1 x））y == \x y - > test2（test1 x）y
I am trying to compose a function of type (Floating a) => a -> a -> a with a function of type (Floating a) => a -> a to obtain a function of type (Floating a) => a -> a -> a. I have the following code:
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)However, when I compile it in GHCI, I get the following error:
/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.Note that the commented-out version of testBoth compiles. The strange thing is that if I remove the (Floating a) constraints from all type signatures or if I change test1 to just take x instead of x and y, testBoth compiles.
I've searched StackOverflow, Haskell wikis, Google, etc. and not found anything about a restriction on function composition relevant to this particular situation. Does anyone know why this is happening?
解决方案 \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 .) . test1These two things are not like each other.
test2 . test1 == \x -> (test2 . test1) x == \x -> test2 (test1 x) == \x y -> (test2 (test1 x)) y == \x y -> test2 (test1 x) y
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