无法理解函数体内F#可变变量的逻辑 [英] Can't understand the logic of F# mutable variable inside function body

问题描述

我正在学习F#，并陷入了mutable关键字的概念.

I'm learning F# and get stuck with the concept of mutable keyword.

请参见以下示例:

let count =
    let mutable a = 1
    fun () -> a <- a + 1; a

val count: unit -> int

每次用()调用时，该值增加1.但是下一个代码没有:

Which increases by 1 every time it's called with (). But next code does not:

let count =
    let mutable a = 1
    a <- a + 1
    a

val count: int

始终是2.

在我正在学习的书中，它用第一个示例说:当函数第一次调用时，可变值a的初始化仅进行一次."

In the book I'm studying with, it says with the first example, "The initialization of mutable value a is done only once, when the function has called first time."

当我开始使用haskell学习FP时，它处理这种副作用的方式完全烧伤了我的大脑，但是F#mutable却以另一种方式再次破坏了我的大脑.以上两个摘要有什么区别?而且，关于可变值的初始化，以上句子的真正含义和条件是什么?

When I started learning FP with haskell, the way it handled side effects like this totally burnt my brain, but F# mutable is destroying my brain again, with a different way. What's the difference between above two snippets? And, what's the true meaning and condition of above sentence, about the initialization of mutable value?

推荐答案

第二个示例

let count =
    let mutable a = 1
    a <- a + 1
    a

定义一个初始化为1的可变变量，然后使用<-运算符为其分配新值(a + 1)，然后在最后一行返回更新的值.由于a具有类型int，并且该类型是从函数返回的，因此该函数的返回类型也是int.

Defines a mutable variable initialised to 1, then assigns a new value (a + 1) to it using the <- operator before returning the updated value on the last line. Since a has type int and this is returned from the function the return type of the function is also int.

第一个例子

let count =
    let mutable a = 1
    fun () -> a <- a + 1; a

还声明了一个初始化为1的int.但是，不是直接返回它，而是返回在a上关闭的函数.每次调用此函数时，a都会递增，并返回更新的值.它可以等效地写为:

also declares an int a initialised to 1. However instead of returning it directly it returns a function which closes over a. Each time this function is called, a is incremented and the updated value returned. It could be equivalently written as:

let count =
    let mutable a = 1
    let update () =
        a <- a + 1
        a
    update

fun () -> ...定义了

fun () -> ... defines a lambda expression. This version returns a 1-argument function reflected in the different return type of unit -> int.

这篇关于无法理解函数体内F#可变变量的逻辑的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！