尾调用

由此可见，"尾调用优化"对递归操作意义重大，所以一些函数式编程语言将其写入了语言规格。ES6也是如此，第一次明确规定，所有 ECMAScript 的实现，都必须部署"尾调用优化"。这就是说，在 ES6 中，只要使用尾递归，就不会发生栈溢出，相对节省内存。

四、递归函数的改写

尾递归的实现，往往需要改写递归函数，确保最后一步只调用自身。做到这一点的方法，就是把所有用到的内部变量改写成函数的参数。比如上面的例子，阶乘函数 factorial 需要用到一个中间变量 total ，那就把这个中间变量改写成函数的参数。这样做的缺点就是不太直观，第一眼很难看出来，为什么计算5的阶乘，需要传入两个参数5和1？

两个方法可以解决这个问题。方法一是在尾递归函数之外，再提供一个正常形式的函数。

function tailFactorial(n, total) {

 if(n ===1)

    return total;

 return tailFactorial(n -1, n * total);

}

function factorial(n) {

  return tailFactorial(n,1);

}

factorial(5) // 120

上面代码通过一个正常形式的阶乘函数 factorial ，调用尾递归函数 tailFactorial ，看起来就正常多了。

函数式编程有一个概念，叫做柯里化（currying），意思是将多参数的函数转换成单参数的形式。这里也可以使用柯里化。

function currying(fn, n) {

  return function(m){

           return fn.call(this, m, n);

         };
}

function tailFactorial(n, total) {

  if(n ===1)

      return total;

  return tailFactorial(n -1, n * total);

}

const factorial = currying(tailFactorial,1);

factorial(5) // 120

上面代码通过柯里化，将尾递归函数 tailFactorial 变为只接受1个参数的 factorial 。

第二种方法就简单多了，就是采用ES6的函数默认值。

function factorial(n, total =1) {if(n ===1)return total;returnfactorial(n -1, n * total);}

factorial(5) // 120

上面代码中，参数 total 有默认值1，所以调用时不用提供这个值。

总结一下，递归本质上是一种循环操作。纯粹的函数式编程语言没有循环操作命令，所有的循环都用递归实现，这就是为什么尾递归对这些语言极其重要。对于其他支持"尾调用优化"的语言（比如Lua，ES6），只需要知道循环可以用递归代替，而一旦使用递归，就最好使用尾递归。

Tail-call optimization is where you are able to avoid allocating a new stack frame for a function because the calling function will simply return the value that it gets from the called function. The most common use is tail-recursion, where a recursive function written to take advantage of tail-call optimization can use constant stack space.

Scheme is one of the few programming languages that guarantee in the spec that any implementation must provide this optimization (JavaScript will also, once ES6 is finalized), so here are two examples of the factorial function in Scheme:

(define(fact x)(if(= x 0)1(* x (fact (- x 1)))))(define(fact x)(define(fact-tail x accum)(if(= x 0) accum
        (fact-tail (- x 1)(* x accum))))(fact-tail x 1))

The first function is not tail recursive because when the recursive call is made, the function needs to keep track of the multiplication it needs to do with the result after the call returns. As such, the stack looks as follows:

(fact 3)(*3(fact 2))(*3(*2(fact 1)))(*3(*2(*1(fact 0))))(*3(*2(*11)))(*3(*21))(*32)6

In contrast, the stack trace for the tail recursive factorial looks as follows:

(fact 3)(fact-tail 31)(fact-tail 23)(fact-tail 16)(fact-tail 06)6

As you can see, we only need to keep track of the same amount of data for every call to fact-tail because we are simply returning the value we get right through to the top. This means that even if I were to call (fact 1000000), I need only the same amount of space as (fact 3). This is not the case with the non-tail-recursive fact, and as such large values may cause a stack overflow.