Haskell使用什么类型的范围? [英] What type of scope does Haskell use?
问题描述
我试图弄清楚Haskell是使用动态还是静态作用域。
我意识到,例如,如果您定义:
let x = 10
然后定义函数
let square 2 = x * x
你有两个不同的x's,这是否意味着它是动态范围?如果没有,它使用的范围是什么,为什么?
另外,Haskell变量是否有别名(同一个内存位置/值的名称不同)?
谢谢。
- Haskell中没有可变变量只是定义(或不可变变量)
- 变量存储器location是一个在Haskell中不存在的概念
在您的示例中, x 不是不是在这个例子中,函数中的10只是square的一个参数,它可以取任何值(你可以指定类型),但在这种情况下。
以下是由 Curt Sampson 提供的别名示例:
import Data.IORef
main :: IO()
main = do x < - newIORef 0 - 将0写入x
readIO Ref x>> = print - x包含0
let y = x
readIORef y>> = print - y包含0
writeIORef x 42 - x
readIORef y>> = print - y包含42
I'm trying to figure out if Haskell uses dynamic or static scoping. I realize that, for example, if you define:
let x = 10
then define the function
let square x = x*x
You have 2 different "x's", and does that mean it is dynamically scoped? If not, what scoping does it use, and why?
Also, can Haskell variables have aliases (a different name for the same memory location/value)?
Thanks.
There are some things wrong in your statements...
- There are no mutable variables in Haskell just definitions (or immutable variables)
- A variable memory location is a concept that do not exist in Haskell
In your example, x is not 10 in the function is just a argument to square, that can take any value (you can specify the type later) in this case 10 but just in this case.
Here is an example of aliases provided by Curt Sampson:
import Data.IORef
main :: IO ()
main = do x <- newIORef 0 -- write 0 into x
readIORef x >>= print -- x contains 0
let y = x
readIORef y >>= print -- y contains 0
writeIORef x 42 -- write 42 into x
readIORef y >>= print -- y contains 42
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