Question

我正在尝试测试重复列表，但是当我编译并输入

时

repeated [1,2,3,4]

它输出True。怎么了？

belongs :: Eq a => a -> [a] -> Bool
belongs n [] = False
belongs n (x:xs) | n == x = True
                 | otherwise = belongs n xs

repeated :: [Integer] -> Bool
repeated [] = False
repeated (x:xs) | belongs x xs = True
                | otherwise = belongs (head xs) xs

Answer 1

“属于（头部xs）xs”检查xs的头部是否在xs内，这将始终为真。

（除非xs为空，否则程序将崩溃！“ head”是部分函数，使用空列表将崩溃）

这将解决它（也由@talex指出，但我也建议使其更通用，无需将其专门用于Integer）

repeated :: Eq a => [a] -> Bool
repeated [] = False
repeated (x:xs) | belongs x xs = True
                | otherwise = repeated xs

Answer 2

您想要

onChange = (e) => {
 const { name, value } = e.target;
 this.setState({
 [name]: value
  });
};

确定Haskell列表中是否存在重复的元素

2 个答案: