how to write numbers in lambda using ghci

I am new to Haskell, using Ghci.

I have a function, called three, that I want to write as

let three =  \x->(\y->(x(x(x y)))) 

OK, this works, but when I try 开发者_如何转开发

three (2+) 4

It does not work. Instead, I get some "cannot construct infinite type" error.

Please help me.

---

ghci> let three = \x->(\y->(x(x(x y))))
ghci> three (2+) 4
10
ghci> three return "deconstructivist"

<interactive>:1:6:
    Occurs check: cannot construct the infinite type: t = m t
      Expected type: t
      Inferred type: m t
    In the first argument of 'three', namely 'return'
    In the expression: three return "deconstructivist"
ghci> :t three
three :: (t -> t) -> t -> t
ghci> :t return
return :: (Monad m) => a -> m a

• The example you supplied of three (2+) 4, works! Better check that the examples you provide actually reproduce your problem.
• As for with a different example, like the one above with return, the thing is that return results in a different type than the one given. If the type was the same, it would be infinite (and of kind * -> * -> * -> ...), which Haskell does not support.

---

The example you give does work. Let's explain why:

three f = f . f . f
-- so...
three :: (a -> a) -> a -> a

The function needs to have type a -> a because it will receive it's own argument, which requires a type. (2+) has type Num a => a -> a, so three (2+) 4 will work just fine.

However, when you pass a function like return of type Monad m => a -> m a, which returns a different type, it will not match the (a -> a) requirement we set out. This is where and when your function will fail.

While you're at it, try making a function like doTimes with type Integer -> (a -> a) -> a -> a which does the given function the given number of times - it's a good next step after making this function.