Question

我尝试实现一个基本的lambda函数但是遇到了一些错误，并且在搜索这里的问题之后无法找出解决方案。我的代码是：

myMap :: (a -> b) -> [a] -> [b]
myMap addSomething [] = []
myMap addSomething (x:xs) = addSomething x : myMap addSomething xs

-- instance Show (a -> b) where
--          show a = "funcion"

list = [0..4]
listTwo = [(5 :: Int)..9]

addSomething :: Int -> Int
addSomething x = x + 1

addSomethingTwo ::  Num a => a -> a-> a
addSomethingTwo x = (\x->x+1)

main = do
    print $ myMap addSomething list
    print $ myMap addSomethingTwo listTwo

这是我收到的错误消息

 No instance for (Show (Int -> Int)) arising from a use of `print'
    Possible fix: add an instance declaration for (Show (Int -> Int))
    In the expression: print
    In a stmt of a 'do' block: print $ myMap addSomethingTwo listTwo
    In the expression:
      do { print $ myMap addSomething list;
           print $ myMap addSomethingTwo listTwo }

如果我取消注释这些行

instance Show (a -> b) where
         show a = "function"

我得到了这个奇怪的结果

[1,2,3,4,5]
[function,function,function,function,function]
[Finished in 0.4s]

提前谢谢你，的Tamas

Answer 1

addSomethingTwo有一个函数（Int - ＆gt; Int）作为结果，而不是Int

您需要将功能定义为

addSomethingTwo x = (\x->x+1) x

或者，要清除lambda和函数参数中的x是不同的变量（它们在不同的范围内）：

addSomethingTwo y = (\x->x+1) y

Answer 2

class ExampleNotificationOpenedHandler implements OneSignal.NotificationOpenedHandler { // This fires when a notification is opened by tapping on it. @Override public void notificationOpened(OSNotificationOpenResult result) { OSNotificationAction.ActionType actionType = result.action.type; JSONObject data = result.notification.payload.additionalData; String customKey; if (data != null) { customKey = data.optString("customkey", null); if (customKey != null) Log.i("OneSignalExample", "customkey set with value: " + customKey); } if (actionType == OSNotificationAction.ActionType.ActionTaken) Log.i("OneSignalExample", "Button pressed with id: " + result.action.actionID); // The following can be used to open an Activity of your choice. // Replace - getApplicationContext() - with any Android Context. // Intent intent = new Intent(getApplicationContext(), YourActivity.class); // intent.setFlags(Intent.FLAG_ACTIVITY_REORDER_TO_FRONT | Intent.FLAG_ACTIVITY_NEW_TASK); // startActivity(intent); // Add the following to your AndroidManifest.xml to prevent the launching of your main Activity // if you are calling startActivity above. /* <application ...> <meta-data android:name="com.onesignal.NotificationOpened.DEFAULT" android:value="DISABLE" /> </application> /* } }生成一系列函数。它的类型为myMap addSomethingTwo listTwo，因为[Int->Int]是二进制函数addSomethingTwo（提供a->(a->a)），这意味着它需要多个类型Num a并返回数值函数a，感谢currying。这些函数a->a由a->a放入列表中。

默认情况下无法打印功能，因为错误消息报告：＆＃34; map没有Show实例。您评论的Int -> Int实例允许将功能打印为＆＃34; Show＆＃34;。

我对你实际想要实现的目标一无所知。

没有（Show（Int - ＆gt; Int））的实例来自于使用'print＆＃39;

2 个答案: