Tail recursion in NodeJS

问题

So I recently came across case I needed to write code where callback calls itself and so on and wondered about NodeJS and tail-call support, so I found this answer https://stackoverflow.com/a/30369729 saying that yup, it's supported.

So I tried it with this simple code:

"use strict";
function fac(n){
    if(n==1){
        console.trace();
        return 1;
    }
    return n*fac(n-1);
}

fac(5);

Using Node 6.9.2 on Linux x64 and run it as node tailcall.js --harmony --harmony_tailcalls --use-strict and result was:

Trace
    at fac (/home/tailcall.js:4:11)
    at fac (/home/tailcall.js:7:11)
    at fac (/home/tailcall.js:7:11)
    at fac (/home/tailcall.js:7:11)
    at fac (/home/tailcall.js:7:11)
    at Object.<anonymous> (/home/tailcall.js:10:1)
    at Module._compile (module.js:570:32)
    at Object.Module._extensions..js (module.js:579:10)
    at Module.load (module.js:487:32)
    at tryModuleLoad (module.js:446:12)

Which clearly shows callstack gets filled with calls and tail-recursion isn't supported although I use latest NodeJS.

Does NodeJS/JavaScript support tail-recursion at all? Or do I really have to go with generators and yields, but problem here is my callbacks are gonna be heavily asynchronous and I'm not gonna work with return value anyway, I just need to make sure the callstack doesn't get uselessly filled with calls while function refers to itself in return.

回答1:

First off, if your actual case you're concerned about is when a function calls itself from an async callback, then you likely don't have any stack buildup there anyway.

That's because the original function has already returned and the whole stack unwound before the async callback gets called, so though it visually looks like recursion, there is no stack build up.

Here's a simple code example of making multiple sequenced network requests where a function calls itself from an async callback and there is no stack buildup.

function getPages(baseURL, startParam, endParam, progressCallback) {

    function run(param) {
         request(baseURL + param, function(err, response, body) {
             if (err) return progressCallback(err);
             ++param;
             if (param < endParam) {
                 progressCallback(null, body);
                 // run another iteration
                 // there is no stack buildup with this call
                 run(param);
             } else {
                 // indicate completion of all calls
                 progressCallback(null, null);
             }
         });
    }
    run(startParam);
}

The prior invocation of run() has already returned and the stack has completely unwound before the async callback is called so while this visually looks like recursion, there is no stack buildup.

In the specific code you show, you can avoid the recursion entirely by rewriting using a while loop which would work efficiently in any version of Javascript:

function fac(n){
    var total = 1;
    while (n > 1) {
        total *= n--;
    }
    return total;
}

// run demo in snippet
for (var i = 1; i <= 10; i++) {
    console.log(i, fac(i))
}

回答2:

What you have there is not a tail-call. A tail call is a function call performed as the final action of a another function. A tail-recursive call is the same except the function calls itself.

However, your code's final action is n*fac(n-1), not fac(n-1). This is not a recursive tail call because the current stack still needs to remember n while computing the recursive calls so it will know which numbers to multiply.

What you can do is compute this information 1 step before:

const fac = (n, result = 1) =>
  n === 1
    ? result
    : fac(n - 1, n * result);

console.log(fac(5)); // 120

Or in terms of your code:

function fac(n, result = 1) {
  // base case
  if (n === 1) {
    return result;
  }

  // compute the next result in the current stack frame
  const next = n * result;

  // propagate this result through tail-recursive calls
  return fac(n - 1, next);
}

Here's the stacktrace from Chrome Canary:

回答3:

I am not sure your recursive function has a tail call. May be you can try the following;

"use strict";
var fac = (n, r = 1) => n === 1 ? r : (r *= n, fac(n-1,r));
console.log(fac(5));

来源：https://stackoverflow.com/questions/41666751/tail-recursion-in-nodejs

标签

javascript

node.js

recursion

tail-call-optimization