在Haskell中使用'let'功能纯度 [英] Functional Purity using 'let' in Haskell

问题描述

当我正在学习Haskell时，我明白它是一种纯粹的函数式语言。我无法理解为什么让 -statements不违反纯度。

As I am working on learning Haskell, I understand it is a purely functional language. I am having trouble understanding why let-statements don't violate purity.

例如（以ghci）：

For example (in ghci):

Prelude> let e = exp 1
Prelude> e
2.718281828459045
Prelude> let e = 2
Prelude> e
2

不是我的第二个 let 产生副作用的声明？或者是第二个 let 语句新的关闭？

isn't my second let statement producing a side effect? Or is the second let statement a new closure?

推荐答案

让为 e 创建一个新的绑定，该绑定会隐藏现有变量。它不会修改 e 。您可以使用以下方式轻松检查：



Your second let creates a new binding for e that shadows the existing variable. It does not modify e. You can easily check this with the following:

Prelude> let e = 1
Prelude> let f () = "e is now " ++ show e
Prelude> f ()
"e is now 1"
Prelude> let e = 2
Prelude> e
2
Prelude> f ()
"e is now 1"
Prelude> 

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