recursion without explicitly setting variable javascript

Question

Here are two working versions (for Node.js) of the same recursive function.

Version 1

function getDependencies(mod, result) {
  result = result || []
  var dependencies = mod.dependencies || []
  Object.keys(dependencies).forEach(function(dep) {
    var key = dep + '@' + mod.dependencies[dep].version
    if (result.indexOf(key) === -1) result.push(key)
    getDependencies(mod.dependencies[dep], result) // COMPARE THIS LINE
  })
  return result.sort()
}

Version 2

function getDependencies(mod, result) {
  result = result || []
  var dependencies = mod.dependencies || []
  Object.keys(dependencies).forEach(function(dep) {
    var key = dep + '@' + mod.dependencies[dep].version
    if (result.indexOf(key) === -1) result.push(key)
    result = getDependencies(mod.dependencies[dep], result) // COMPARE THIS LINE
  })
  return result.sort()
}

How does version 1 of the function work, as compared to version 2, without explicitly setting the result variable?

Is your semi-colon key broken?

jfriend00
– jfriend00

2014-11-02 06:17:06 +00:00
Commented Nov 2, 2014 at 6:17 — jfriend00
– jfriend00, Commented Nov 2, 2014 at 6:17
This is a Node.js solution from Nodeschool.

user3477405
– user3477405

2014-11-02 06:17:58 +00:00
Commented Nov 2, 2014 at 6:17 — user3477405
– user3477405, Commented Nov 2, 2014 at 6:17

Barmar · Accepted Answer · 2014-11-02 06:18:11Z

1

result.push(key) and result.sort() both modify the result array that was passed in as an argument. So there's no need to assign it again in the caller.

answered Nov 2, 2014 at 6:18

Barmar

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3 Comments

user3477405 Over a year ago

So is it changed because of pass by reference? If getDepenencies() is recursively called multiple times, I do not understand how the results "add up" and are finally returned as one value.

Barmar Over a year ago

When you pass or assign an object or array, you get a reference to the same object/array, it doesn't make a copy. So all the recursive calls are operating on the same result array.

jfriend00 Over a year ago

@user3477405 - there's only one array. When you pass it to a function, no copy is made - it's the same array. So, if a function modifies it, the original is modified. No need to return it. It isn't technically pass by reference, more like pass by pointer.

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