Haskell有什么样的'约束'意思 [英] What does has kind 'Constraint' mean in Haskell
问题描述
我试图做的功能是解决任意维数向量的毕达哥拉斯定理。这看起来像这样：平方根（a ^ 2 + b ^ 2 + c ^ 2 ...）
我可以在ghci中做几行操作，我试图做一个函数如下：
pre $ code sq x = x * x
b = map sq [1,2,3]
a = sum b
x = sqrt b
当我这样做时，我试图包含许多类型的签名，
目前我的函数如下所示：
mod :: [Num a] => a
mod x = sqrt a
其中a = sum [b]
其中[b] = map sq [x]
当我尝试运行它时，我不明白这个问题：
预期约束，但'[Num a]'有'*'
•在类型签名中：
Main.mod :: [Num a] => a
几件事要调整：
0） mod 对您的函数来说不是一个好名字，因为它是标准库中模函数的名称。我将它称为 norm 。
1）您打算编写的类型签名是：
norm :: Num a => [a] - > a
[a] 是一个元素类型为 a 的列表。 => 之前的 Num a 不是类型，而是约束 ，它指定 a 必须是一个数字类型（或者更准确的说，它必须是 Num 类）。 [Num a] => 会导致您所看到的错误，因为在给定方括号的情况下，类型检查器将其视为尝试使用列表类型而不是约束。
除了 Num a 问题之外，您从签名中省略了结果类型。校正后的签名反映出你的函数需要一个数字列表并返回一个数字。
<2>
Num a 对于你想要做的事情，约束太弱。为了使用 sqrt ，您不仅需要一个数字类型，而且还需要一个浮动 （参考左边的评论对这个答案）： GHCi> ：t sqrt
sqrt :: Floating a => a - > a
因此，您的签名应该是
norm :: Floating a => [a] - > a
<3> [x] 是一个列表中包含一个元素 x 。如果你的参数已经是一个列表，就像类型签名所说的那样，不需要把它放在方括号中。然后你的函数变成：
norm :: Floating a => [a] - > a
标准x = sqrt a
其中a =总和b
其中b =地图平方x
或者，更整齐地说，没有第二个其中 -block：
norm :: Floating a => [a] - > a
norm x = sqrt（b）
其中b = map sq x
I am fresh to Haskell and I am trying to understand the language by writing some code. I am only familiar with very simple instructions on ghci: head, tail, sum, (*), and the like – very simple.
The function I am trying to make is for solving Pythagoras's theorem for vectors of any number of dimensions. This looks something like this: square root (a^2 + b^2 + c^2 ...)
What I can do in ghci in a few lines, which I am trying to make a function is the following:
sq x = x*x
b = map sq [1,2,3]
a = sum b
x = sqrt b
When I do this I try to include a signature of many sorts, Currently my function looks like this:
mod :: [Num a] => a
mod x = sqrt a
where a = sum [b]
where [b] = map sq [x]
I do not understand the issue when I try to run it:
Expected a constraint, but ‘[Num a]’ has kind ‘*’
• In the type signature:
Main.mod :: [Num a] => a
A few things to adjust:
0) mod isn't a good name for your function, as it is the name of the modulo function from the standard library. I will call it norm instead.
1) The type signature you meant to write is:
norm :: Num a => [a] -> a
[a] is the type of a list with elements of type a. The Num a before the => isn't a type, but a constraint, which specifies that a must be a number type (or, more accurately, that it has to be an instance of the Num class). [Num a] => leads to the error you have seen because, given the square brackets, the type checker takes it as an attempt to use a list type instead of a constraint.
Beyond the Num a issue, you have left out the result type from the signature. The corrected signature reflects that your function takes a list of numbers and returns a number.
2) The Num a constraint is too weak for what you are trying to do. In order to use sqrt, you need to have not merely a number type, but one that is an instance of Floating (cf. leftaroundabout's comment to this answer):
GHCi> :t sqrt
sqrt :: Floating a => a -> a
Therefore, your signature should be
norm :: Floating a => [a] -> a
3) [x] is a list with a single element, x. If your argument is already a list, as the type signature says, there is no need to enclose it in square brackets. Your function, then, becomes:
norm :: Floating a => [a] -> a
norm x = sqrt a
where a = sum b
where b = map sq x
Or, more neatly, without the second where-block:
norm :: Floating a => [a] -> a
norm x = sqrt (sum b)
where b = map sq x
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