编程haskell代码:素数列表 [英] programming haskell code: List of primes

问题描述

我的任务是创建素数列表，但仅包含以下信息:

My task is to create a list of primes but only with the following information:

-列表必须是无限的
-我必须检查n > 1
的素数 -如果有一个变量2 <= k < n-2将其除以n，则它不是素数；如果它没有除以n，则它为
-没有功能！

-The list has to be infinite
-I have to check up primes for n > 1
-If there is a variable 2 <= k < n-2, which divides n,then it is no prime, if it does not divide n, it is
-No function!

所以我开始写这样的代码:

So I started writing a code like this:

primes = [n| n<-[2,3..],k<-[2,3..], if (k>=2) && (k<=(n-2))
            then if n `mod` k /= 0 then n else return ()
            else return ()  ]

但是随后出现以下错误:无法将期望的类型'Bool'与实际类型相匹配"

But then appears the following error : "Couldn't match expected type ‘Bool’ with actual type"

是因为return ()，但是我不知道如何替换此return.

It's because of the return (), but I don't know how to replace this return.

我希望得到帮助.

推荐答案

假设您有一个函数:

isPrime :: Int -> Bool

返回素数为True，否则返回False.然后，您可以定义:

which returned True for primes and False otherwise. Then you could define:

primes = [ n | n <- [2..], isPrime n ]

使用功能all，您可以这样编写isPrime:

Using the function all, you could write isPrime like this:

isPrime n = all (\a -> mod n a /= 0) [2..n-1]

如果应用于xs的每个成员的f为True，则

all f xs返回True.因此，如果对于[2..n-1]中的所有元素mod n a /= 0，即不除n，则n是素数.

all f xs returns True if f applied to every member of xs is True. So n is prime if for all elements in [2..n-1], mod n a /= 0, i.e. a does not divide n.

all函数的更多示例:

all even [2,4,10,12]       -- True
all (\x -> x > 0) [4,-5,6] -- False
all (\xs -> length xs > 0) [ [3], [4,5], [-4] ] -- True
all (\x -> x*x < 0) []     -- True

您可以使用用于列表递归的标准模板自己定义all:

You can come up with the definition of all yourself by using the standard template for recursion on lists:

all f []     =  ...
all f (x:xs) =  ...

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