所以我想制作一个打印出 ASCII 三角形的程序。程序会询问您希望三角形的高度并将其打印出来,直到您输入的数字为7为止,然后绘制高度为 7 的三角形,程序停止。如果数字是偶数,则打印出的三角形由“*”组成,如果数字是奇数,则由“#”组成。我想通过递归调用main函数来做到这一点
triangle level character = draw level character 1
where draw level character accumulator
| level <=0 = putStr("")
| level > 0 = do
putStrLn (replicate level ' ' ++ replicate accumulator character)
draw (level-1) character (accumulator+2)
main = do
number <- readLn :: IO Int
if(number `mod` 2 == 0)
then
triangle number'*'
main
else if (number == 7)
then triangle number '#'
else triangle number '#'
main
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我得到的错误是
• Couldn't match expected type ‘IO () -> IO ()’
with actual type ‘IO ()’
• The function ‘triangle’ is applied to three arguments,
but its type ‘Int -> Char -> IO ()’ has only two
In the expression: triangle number '*' main
In a stmt of a 'do' block:
if (number `mod` 2 == 0) then
triangle number '*' main
else
if (number == 7) then
triangle number '#'
else
triangle number '#' main
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错误说:
• The function ‘triangle’ is applied to three arguments,
but its type ‘Int -> Char -> IO ()’ has only two
In the expression: triangle number '*' main
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所以编译器已将您的代码解析为triangle number '*' main; 这意味着triangle number '*'和之间的换行符main不会被解释为do块语句中断,原因是您不在do块内!同样的错误出现在后面的elsefor 子句中triangle number '#' main。解决方法是添加一个do:
main = do
number <- readLn :: IO Int
if(number `mod` 2 == 0)
then do
triangle number'*'
main
else if (number == 7)
then triangle number '#'
else do
triangle number '#'
mainRun Code Online (Sandbox Code Playgroud)
如果你对 Haskell 的布局规则有疑问,你总是可以使用显式花括号{......}围绕块和;块项之间的分号,以检查你对代码是如何解析的理解。
main :: IO (); -- end signature
main = do { -- begin ‘do’
number <- readLn :: IO Int; -- end statement
if (number `mod` 2 == 0)
then do {
triangle number '*';
main;
}
else if (number == 7)
then triangle number '#'
else do {
triangle number '#';
main;
}; -- end ‘do’, end statement (starting with ‘if…’)
}; -- end ‘do’, end equationRun Code Online (Sandbox Code Playgroud)
另外,请注意,我在字符文字周围放置了空格,… number '*' …而不是… number'*' …. 标识符中允许使用撇号(作为“素数”符号的 ASCII 近似值 ? ),因此如果您将其更改为 ,number'X'那么它将被解析为单个名称number'X',而不是名称number后跟字符文字'X',这可能会造成混淆。
首先,您在do这里缺少一个:
then
triangle number'*'
main
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后面的两行then被一起解释为triangle number '*' main(如果您注意到,这是错误消息中引用的方式)。为了将这两行分别解释为后续操作,它们需要在 a 中do:
then do
triangle number'*'
main
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同样的问题,加上模棱两可的缩进,存在于最后一个块中。大概应该是:
else do
triangle number '#'
main
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