在F#中将递归函数标记为rec的原因是什么？

发布于 2024-09-24 02:41:11 字数 288 浏览 6 评论 0原文

我不确定这是否是一个愚蠢的问题，但我正在浏览 VS 2010 附带的教程，并且有一个这样的函数：

let rec factorial n = if n=0 then 1 else n * factorial (n-1)

What's the Reason of this recursive function to be marked with the rec keywords?

是否可以让编译器确信它是递归的，以便可以进行某些优化？

如果排除它会发生什么？

原文

I am not sure if this is a stupid question but I was going through the tutorial that comes with VS 2010 and there is a function like this:

let rec factorial n = if n=0 then 1 else n * factorial (n-1)

What's the reason of this recursive function to be marked with the rec keyword?

Is it so that the compiler is assured of it being recursive so can do certain optimizations?

What happens if you exclude it?

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此岸叶落 2024-10-01 02:41:11

这可能具有启发性：

let Main() =
    let f(x) = 
        printfn "original f: %d" x
    let f(x) =
    //let rec f(x) =
        printfn "entered new f: %d" x
        if x > 0 then
            f(x-1)
        else
            printfn "done"
    f(3)
Main()

则会打印

entered new f: 3
original f: 2

如果我们注释掉 let 并取消注释 let rec，

entered new f: 3
entered new f: 2
entered new f: 1
entered new f: 0
done

Now，然后它会打印So 从这个角度来看，它只是关于名称绑定； let rec 立即将标识符放入作用域中（在本例中，隐藏了前面的 f），而 let 仅将标识符放入作用域中体被定义。

该规则的动机确实源于与类型推断的交互。

This might be instructive:

let Main() =
    let f(x) = 
        printfn "original f: %d" x
    let f(x) =
    //let rec f(x) =
        printfn "entered new f: %d" x
        if x > 0 then
            f(x-1)
        else
            printfn "done"
    f(3)
Main()

That prints

entered new f: 3
original f: 2

Now if we comment out let and uncomment let rec, then it prints

entered new f: 3
entered new f: 2
entered new f: 1
entered new f: 0
done

So from that point of view, it's just about name binding; let rec puts the identifier in scope immediately (in this example, shadowing the previous f), whereas let puts the identifier in scope only after its body is defined.

The motivation for the rule does stem from interactions with type inference.

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