Haskell 歧义类型

发布于 2024-10-16 21:15:23 字数 766 浏览 9 评论 0原文

findMult lst n = [x | x <- lst, x `mod` n == 0]

primes num = 
    let n = [2..num]
        x = ceiling (sqrt num)
        nsqrt = [2..x]
        not_prime = map (findMult n) nsqrt
    in diff2 n (concat not_prime)

当我尝试运行它时出现以下问题

<interactive>:1:0:
    Ambiguous type variable `t' in the constraints:
      `RealFrac t' arising from a use of `primes' at <interactive>:1:0-8
      `Floating t' arising from a use of `primes' at <interactive>:1:0-8
      `Integral t' arising from a use of `primes' at <interactive>:1:0-8
    Probable fix: add a type signature that fixes these type variable(s)

我尝试使用 fromIntegral 但我认为我使用不正确，因为这给了我编译错误。请帮忙。

这样做的目的是找到 num 之前的所有素数。

原文

findMult lst n = [x | x <- lst, x `mod` n == 0]

primes num = 
    let n = [2..num]
        x = ceiling (sqrt num)
        nsqrt = [2..x]
        not_prime = map (findMult n) nsqrt
    in diff2 n (concat not_prime)

has the following problem when i try to run it

<interactive>:1:0:
    Ambiguous type variable `t' in the constraints:
      `RealFrac t' arising from a use of `primes' at <interactive>:1:0-8
      `Floating t' arising from a use of `primes' at <interactive>:1:0-8
      `Integral t' arising from a use of `primes' at <interactive>:1:0-8
    Probable fix: add a type signature that fixes these type variable(s)

I tried using fromIntegral but i don't think i used correctly as that gives me compilation error. Please help.

The purpose of this is to find all the prime numbers up until num.

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怪我入戏太深 2024-10-23 21:15:23

当您在需要浮点值的地方使用整数值时，您会收到类似这样的错误消息（反之亦然）。

在这种情况下，问题在于您在 num 上调用 sqrt（它采用浮点值作为参数），使编译器认为 num > 是浮点值。但也可以使用 num 作为 n 的上限，它是一个整数值列表（因为它用作 findMult 的参数，需要整数值列表）。

因此，在对 num 调用 sqrt 之前，先调用 fromIntegral ，如下所示：

x = ceiling (sqrt (fromIntegral num))

You get error messages like this when you use an integral value where a floating value was expected (or vice versa).

In this case the problem is that you're calling sqrt, which takes a floating point value as an argument, on num making the compiler think num is a floating point value. But also use num as an upper limit for n, which is a list of integral values (because it's used as an argument to findMult which needs a list of integral values).

So before calling sqrt on num call fromIntegral on it, like this:

x = ceiling (sqrt (fromIntegral num))

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永不分离 2024-10-23 21:15:23

您可以不求平方根，而是求所有平方直到极限。

-- instead of
{- x = ceiling (sqrt num)
   nsqrt = [2..x] -}
-- avoid sqrt
nsqrt = takeWhile (\x -> (x-1)^2 < num) [2..]
-- or even avoid multiplication altogether
nsqrt = map fst . takeWhile ((< num) . snd) . zip [2..] $ scanl1 (+) [1,3..]

Instead of taking the square root, you can take all squares up to the limit.

-- instead of
{- x = ceiling (sqrt num)
   nsqrt = [2..x] -}
-- avoid sqrt
nsqrt = takeWhile (\x -> (x-1)^2 < num) [2..]
-- or even avoid multiplication altogether
nsqrt = map fst . takeWhile ((< num) . snd) . zip [2..] $ scanl1 (+) [1,3..]

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