haskell - 奇怪的 Haskell 类型错误仅在使用 assertEqual 时发生

Question

此代码将正确编译:

import Text.Printf
import Test.HUnit
doubleMe x = x + x
doubleUs x y = doubleMe x + doubleMe y
doubleSmallNumber x = if x > 100 then x else x*2
doubleSmallNumber' x = if x > 100 then x else x*2 + 1
conanO'Brien = "It's a-me, Conan O'Brien!"
main = do
  runTestTT $ TestList [TestCase $ ae 4 $ doubleMe 2,
                        TestCase $ ae 10 $ doubleUs 2 3,
                        TestCase $ ae 4 $ doubleSmallNumber 2,
                        TestCase $ ae 1000 $ doubleSmallNumber' 1000,
                        TestCase $ assertEqual "" "It's a-me, Conan O'Brien!" conanO'Brien]
  where ae = assertEqual "" 

输出是:

$ clear && ghc baby.hs && ./baby     
[1 of 1] Compiling Main             ( baby.hs, baby.o )
Linking baby ...
ld: warning: could not create compact unwind for .LFB3: non-standard register 5 being saved in prolog
Cases: 5  Tried: 5  Errors: 0  Failures: 0

当我将代码更改为:

import Text.Printf
import Test.HUnit
doubleMe x = x + x
doubleUs x y = doubleMe x + doubleMe y
doubleSmallNumber x = if x > 100 then x else x*2
doubleSmallNumber' x = if x > 100 then x else x*2 + 1
conanO'Brien = "It's a-me, Conan O'Brien!"
main = do
  runTestTT $ TestList [TestCase $ ae 4 $ doubleMe 2,
                        TestCase $ ae 10 $ doubleUs 2 3,
                        TestCase $ ae 4 $ doubleSmallNumber 2,
                        TestCase $ ae 1000 $ doubleSmallNumber' 1000,
                        TestCase $ ae "It's a-me, Conan O'Brien!" conanO'Brien]
  where ae = assertEqual "" 

我得到:

[1 of 1] Compiling Main             ( baby.hs, baby.o )

baby.hs:12:65:
    No instance for (Num [Char])
      arising from the literal `1000'
    Possible fix: add an instance declaration for (Num [Char])
    In the first argument of `doubleSmallNumber'', namely `1000'
    In the second argument of `($)', namely `doubleSmallNumber' 1000'
    In the second argument of `($)', namely
      `ae 1000 $ doubleSmallNumber' 1000'

我不明白为什么。

也有人对修复 ld 警告有任何想法:

ld: warning: could not create compact unwind for .LFB3: non-standard register 5 being saved in prolog

Answer 1

这是单态限制的一个例子。 ae “看起来像一个值”(没有参数)并且没有显式类型，因此编译器不会为其推断多态类型。

在第一个示例中，它的类型为 Int -> Int -> Assertion (我认为)。

第二，来自ae "It's a-me, Conan O'Brien!" conanO'Brien它的类型为 String -> String -> Assertion .请记住，整数文字的类型实际上是 Num a => a , 和 1000获取类型 String来自 ae , 所以编译器需要一个实例 Num String .

已编辑:这可以通过给出明确的类型注释来解决:where ae :: (Show a, Eq a) => a -> a -> Assertion = assertEqual "" .或者通过在定义中添加参数(eta-expanding):where ae x y = assertEqual "" x y .