sumOfSquare :: Int -> Int -> Int
sumOfSquare a b = a * a + b * b
hipotenuse :: Int -> Int -> Int
hipotenuse a b = truncate(sqrt(x))
where x = fromIntegral(sumOfSquare a b)
squareCheck :: Int -> Bool
squareCheck n = truncate(sqrt(x)) * truncate(sqrt(x)) == n
where x = fromIntegral n
isItSquare :: Int -> Int -> Bool
isItSquare a b = squareCheck (sumOfSquare a b)
calc :: (Integral a) => a -> [(a, a, a)]
calc a = [(x, y, (hipotenuse x y)) | x <- [1..a], y <-[1..a], (isItSquare x y)]
Error message:
Prelude> :load "some.hs"
[1 of 1] Compiling Main ( some.hs, interpreted )
some.hs:16:74:
Couldn't match expected type `Int' against inferred type `a'
`a' is a rigid type variable bound by
the type signature for `calc' at some.hs:15:18
In the first argument of `isItSquare', namely `x'
In the expression: (isItSquare x y)
In a stmt of a list comprehension: (isItSquare x y)
Failed, modules loaded: none.
As I understand the type of 'x' and 'y'. Is it right? Is it square require the Int. But what is the type 'x' and 'y'? I thinked they are Int.
Your type is too general. You are passing x
and y
to isItSquare
, which is expecing Int
s, but you don't know that x
and y
are Int
s. They could be, but they could be any other instance of Integral
as well. Either change the signature to the more specific:
calc :: Int -> [(Int, Int, Int)]
Or have your helper functions work on more general types:
squareCheck :: (Integral a) => a -> Bool
...
You've declared sumOfSquare
, hipotenuse
, squareCheck
and isItSquare
as operating on Int
.
However, you've said that calc
can use any type a
, as long as a
is Integral
.
Either declare calc
like this:
calc :: Int -> [(Int, Int, Int)]
...or change all your other functions like this:
sumOfSquare :: (Integral a) => a -> a -> a
calc :: (Integral a) => a -> [(a, a, a)]
calc a = [(x, y, (hipotenuse x y)) | x <- [1..a], y <-[1..a], (isItSquare x y)]
a
has type a
(the signature explicitly says so, that's what "is a rigid type variable bound by the type signature for calc
" means), and x
is taken from the list [1..a]
, so it also has type a
(and same for y
).
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