[javascript] Calling a javascript function recursively

I can create a recursive function in a variable like so:

/* Count down to 0 recursively.
 */
var functionHolder = function (counter) {
    output(counter);
    if (counter > 0) {
        functionHolder(counter-1);
    }
}

With this, functionHolder(3); would output 3 2 1 0. Let's say I did the following:

var copyFunction = functionHolder;

copyFunction(3); would output 3 2 1 0 as above. If I then changed functionHolder as follows:

functionHolder = function(whatever) {
    output("Stop counting!");

Then functionHolder(3); would give Stop counting!, as expected.

copyFunction(3); now gives 3 Stop counting! as it refers to functionHolder, not the function (which it itself points to). This could be desirable in some circumstances, but is there a way to write the function so that it calls itself rather than the variable that holds it?

That is, is it possible to change only the line functionHolder(counter-1); so that going through all these steps still gives 3 2 1 0 when we call copyFunction(3);? I tried this(counter-1); but that gives me the error this is not a function.

You can access the function itself using arguments.callee ^[MDN]:

if (counter>0) {
    arguments.callee(counter-1);
}

This will break in strict mode, however.

Here's one very simple example:

var counter = 0;

function getSlug(tokens) {
    var slug = '';

    if (!!tokens.length) {
        slug = tokens.shift();
        slug = slug.toLowerCase();
        slug += getSlug(tokens);

        counter += 1;
        console.log('THE SLUG ELEMENT IS: %s, counter is: %s', slug, counter);
    }

    return slug;
}

var mySlug = getSlug(['This', 'Is', 'My', 'Slug']);
console.log('THE SLUG IS: %s', mySlug);

Notice that the counter counts "backwards" in regards to what slug's value is. This is because of the position at which we are logging these values, as the function recurs before logging -- so, we essentially keep nesting deeper and deeper into the call-stack before logging takes place.

Once the recursion meets the final call-stack item, it trampolines "out" of the function calls, whereas, the first increment of counter occurs inside of the last nested call.

I know this is not a "fix" on the Questioner's code, but given the title I thought I'd generically exemplify Recursion for a better understanding of recursion, outright.

You can use the Y-combinator: (Wikipedia)

// ES5 syntax
var Y = function Y(a) {
  return (function (a) {
    return a(a);
  })(function (b) {
    return a(function (a) {
      return b(b)(a);
    });
  });
};

// ES6 syntax
const Y = a=>(a=>a(a))(b=>a(a=>b(b)(a)));

// If the function accepts more than one parameter:
const Y = a=>(a=>a(a))(b=>a((...a)=>b(b)(...a)));

And you can use it as this:

// ES5
var fn = Y(function(fn) {
  return function(counter) {
    console.log(counter);
    if (counter > 0) {
      fn(counter - 1);
    }
  }
});

// ES6
const fn = Y(fn => counter => {
  console.log(counter);
  if (counter > 0) {
    fn(counter - 1);
  }
});

I know this is an old question, but I thought I'd present one more solution that could be used if you'd like to avoid using named function expressions. (Not saying you should or should not avoid them, just presenting another solution)

  var fn = (function() {
    var innerFn = function(counter) {
      console.log(counter);

      if(counter > 0) {
        innerFn(counter-1);
      }
    };

    return innerFn;
  })();

  console.log("running fn");
  fn(3);

  var copyFn = fn;

  console.log("running copyFn");
  copyFn(3);

  fn = function() { console.log("done"); };

  console.log("fn after reassignment");
  fn(3);

  console.log("copyFn after reassignment of fn");
  copyFn(3);