小编Wal*_*lle的帖子

我正在编写一个函数来检查树是否是BST.我所尝试的只是按顺序遍历列表打印树,然后检查列表是否在增加.但是我遇到了这个错误:

Couldn't match expected type `a' against inferred type `[t]'
  `a' is a rigid type variable bound by
      the type signature for `checkList' at BST.hs:24:18
In the pattern: x : y : xs
In the pattern: [x : y : xs]
In the definition of `checkList':
    checkList [x : y : xs] = x <= y && checkList (y : xs)

这是我到目前为止(只有一个checkList函数).

checkList :: (Ord a) => [a] -> Bool
checkList [] = True
checkList [x] = True
checkList [x:y:xs] …

我正在测试一个小程序,它基本上比较了2个输入字符串是否相同(就像strcmp那样).然而,我不断得到结果说2串不管是什么.如果有人能发现一些错误,我会很感激

int comp(char str1[], char str2[]) {
    int i = 0;
    while (str1[i] == str2[i]) {
        if (str1[i] == '\0' || str2[i] == '\0')
            break;
        i++;
    }
    if (str1[i] == '\0' && str2[i] == '\0')
        return 0;
    else
        return -1;
}


int main(int argc, char * * argv) {
    int cmp;
    char str1[1000], str2[1000];
    cmp = comp(str1, str2);
    if (cmp == 0)
        printf("The two strings are identical.\n");
    else
        printf("The two strings are different.\n");
    return 0;
}

假设我有一个BST的数据类型:

data tree a = Empty | Node a (tree a) (tree a)
                    deriving (Show, Read, Eq)

我正在做一个简单的map函数来应用BST的每个元素.

treeMap :: (a -> b) -> tree a -> tree b
treeMap f (Empty) = Empty
treeMap f (Node left right) = Node (treeMap f left) (treeMap f right)

但它给我一个错误说:

Constructor `Node' should have 3 arguments, but has been given 2
In the pattern: Node left right
In the definition of `treeMap':
    treeMap f (Node left right)
              = Node (treeMap f left) …

我正在测试一个小程序,它基本上比较了2个输入字符串是否相同(就像strcmp那样).

我想做类似的事情(用户在同一行输入2个所需的字符串).在这种情况下,它应该返回"两个字符串是不同的"

./a.out foo bar 

我应该这样做来读取用户的输入字符串吗？

 scanf("%s %s", str1, str2);

要么

gets(str1); gets(str2);

这是我到目前为止(由于某些原因似乎陷入无限循环)

int mystrcmp(char str1[], char str2[]) {
    int i = 0;
    while (str1[i] == str2[i]) {
        if (str1[i] == '\0' || str2[i] == '\0') break;
        i++;
    }
    if (str1[i] == '\0' && str2[i] == '\0')
        return 0;
    else
        return -1;
}

int main(int argc, char * * argv) {
    int cmp;
    char str1[1000], str2[1000];
    scanf("%s %s", str1, str2);
    //gets(str1); gets(str2);
    cmp = mystrcmp(str1, str2);
    if (cmp == …