haskell - Haskell中的多个where语句有什么办法吗

Question

我试图在一个函数中编写 3-4 where 语句，但我得到错误并且无法做到，我试图做这样的事情：

foo x=
| x == foo1 = 5
| x == foo2 =3
| x == foo3 =1
| otherwise =2 
where foo1= samplefunct1 x
      foo2= samplefunct2 x
      foo3= samplefunct3 x

我知道代码有点没用，但我写这个只是为了举例说明我的意思。

有没有人可以帮助我？提前致谢。

Answer 1

删除=之后foo x并缩进您的代码，例如

foo x
    | x == foo1 = 5
    | x == foo2 =3
    | x == foo3 =1
    | otherwise =2 
    where foo1 = samplefunct1 x
          foo2 = samplefunct2 x
          foo3 = samplefunct3 x

你很好。

Answer 2

If your indentation is a bit uneven, like this:

foo x
 | x == foo1 = 5
 | x == foo2 =3
 | x == foo3 =1
 | otherwise =2 
 where foo1= samplefunct1 x
        foo2= samplefunct2 x
         foo3= samplefunct3 x

then indeed, the error message talks about unexpected = _{(and in the future, please do include full error message in the question body).}

You fix this error by re-aligning, or with explicit separators { ; }, making it white-space–insensitive:

foo x
 | x == foo1 = 5
 | x == foo2 =3
 | x == foo3 =1
 | otherwise =2 
 where { foo1= samplefunct1 x ;
        foo2= samplefunct2 x ;
          foo3= samplefunct3 x }

This runs fine (not that it is a nice style to use). Sometimes it even looks even to you, but isn't, if there are some tab characters hiding in the white-space.

Answer 3

这段代码几乎是正确的，你只需要正确的缩进：空格在 haskell 中很重要。此外，使用=afterfoo是守卫的错误，因此您也必须将其删除。结果是：

foo x
  | x == foo1 = 5
  | x == foo2 =3
  | x == foo3 =1
  | otherwise =2 
  where foo1= whatever1 x
        foo2= whatever2 x
        foo3= whatever3 x