F#中的增量值 [英] Increment value in F#

问题描述

也许这太简单了，但是我在网上找不到任何答案
我尝试在F#中增加值(如C#中的count++).
我不想使用可变"选项， 我只想看一个例子，F#中的Increment函数应该是什么样子.
以及如何使用它.

Maybe it's too simple thing to do, but I can't find any answer in the web
I'm try to Increment value in F# (like count++ in C#).
I don't want to use "mutable" option, I'm just want to see an example, who Increment function in F# should look like.
And how do I use it.

推荐答案

与C ++相同，递增值"的想法只有在使用可变值或使用变量时才有意义.可变参考单元格(本质上是一个存储可变值的简单对象).如果参考单元可变，则可以使用incr函数:

The idea of "incrementing a value" in the same sense as in C++ only makes sense when you're working with mutable values or when you're using a mutable reference cell (essentially a simple object that stores a mutable value). If you have a mutable reference cell, you can use incr function:

let count = ref 0
incr count

如果使用可变变量，则没有内置函数，您需要编写count + 1:

If you use a mutable variable, then there is no built-in function for this and you need to write count + 1:

let mutable count = 0
count <- count + 1

如果您使用不可变的值编写代码，则通常只需编写count + 1，然后将结果传递给某个函数(或其他函数，具体取决于具体情况).例如，要计算F#列表的长度，应编写:

If you're writing code using immutable values, then you will generally just write count + 1 and then pass the result to some function (or somewhere else - this depends on the specific case). For example, to calculate the length of an F# list, you would write:

let rec length list =
  match list with 
  | [] -> 0
  | _::tail -> 1 + (length tail)

在此示例中，表达式1 + (...)是C ++代码中与i++相对应的代码，该代码在列表上进行迭代并计算其长度.表达式的结果未分配给新变量，因为它是length函数的结果直接返回.

In this example, the expression 1 + (...) is the code corresponding to i++ in a C++ code that iterates over a list and computes its length. The result of the expression is not assigned to a new variable, because it is returned directly as a result of the length function.

编辑函数的参数是不可变的，这意味着您无法更改其值.正如Lee所提到的，您可以使用变量阴影来用新值隐藏旧值-但请注意，这仅具有局部效果(就像定义一个具有不同名称的新变量来存储新值).例如:

EDIT Parameters of functions are immutable meaning that you cannot change their values. As mentioned by Lee, you can use variable shadowing to hide the old value with a new one - but note that this only has a local effect (it is like defining a new variable with different name to store the new value). For example:

let rec length list count =
  match list with 
  | [] -> count
  | _::tail -> 
     let count = count + 1 // Variable shadowing used here
     length tail count

您无法编写函数来简化let count = count + 1行，如上所述，这等效于编写let newCount = count + 1然后在最后一行使用newCount.

You cannot write a function to simplify the line let count = count + 1 and as mentioned above, this is equivalent to writing let newCount = count + 1 and then using newCount on the last line.

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