Javascript function confusion

Question

Firstly, sorry for the not specific title. I don't know how to call this.

Ok, here goes my question.

I am going to take Javascript's reduce function here.

I know the "reduce" function is defined as follows

function reduce(array, combine, start) {
  var current = start;
  for (var i = 0; i < array.length; i++)
    current = combine(current, array[i]);
  return current;
}

Well, what I don't understand is this.

Why

[1, 2, 3, 4].reduce(function(a, b) {
  return a + b;
}, 0);

and

reduce([1, 2, 3, 4], function(a, b) {
  return a + b;
}, 0);

returns the same result of "10"?

I know how the function works. I just don't understand why

xx.functionABC(param2,param3)

is the same as
functionABC(param1,param2,param3)

Thanks for the reply everyone!

Thanks for all the comments and answers!

I feel stupid for not checking with other methods first (or shouldn't try with a built in function).

To extend the comment above. You have defined your own reduce function. But reduce function is also part of the array's prototype. It was just coincidence that both work. — Mouser
– Mouser, Commented Dec 26, 2014 at 12:41
(After your edit) They are not the same, they do the same. This will not work with any random function. — Jongware
– Jongware, Commented Dec 26, 2014 at 12:43
@Jongware, I get it. You should put as answer as I feel thats the best answer so far (I immediately understood what a stupid question this was) — Kiong
– Kiong, Commented Dec 26, 2014 at 12:47

Kendall Frey · Accepted Answer · 2014-12-26 12:49:06Z

2

The first code snippet is using Array.prototype.reduce, and the second snippet is using your custom reduce function.

They happen to be implemented mostly the same, and therefore give the same result.

Here is an example of how they are different. Array.prototype.reduce passes additional arguments to the combine callback, namely the current array index, and a reference to the array itself. Your function does not pass these arguments.

function reduce(array, combine, start) {
  var current = start;
  for (var i = 0; i < array.length; i++)
    current = combine(current, array[i]);
  return current;
}

function foo(a, b, i) { return a + b * i; };

console.log([1, 2, 3, 4].reduce(foo, 0));

console.log(reduce([1, 2, 3, 4], foo, 0));

^{(See the developer console of your browser for the result.)}

edited Dec 26, 2014 at 12:49

answered Dec 26, 2014 at 12:41

Kendall Frey

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2 Comments

Kiong Over a year ago

Thanks a lot. This helped me understand what was wrong with my question. well, after your edit, I don't understand what you meant by "Array.prototype.reduce passes additional arguments to the combine callback, namely the current array index, and a reference to the array itself. Your function does not pass these arguments."

Kendall Frey Over a year ago

Read the reduce documentation, where it says the callback is "Function to execute on each value in the array, taking four arguments". Also see the code to implement it, where it does value = callback(value, t[k], k, t);

Vladimir Bredihin · Accepted Answer · 2014-12-26 12:43:47Z

0

There is has a method Array.prototype.reduce() in js. It means method belongs to the Array.prototype. But there is no simple function reduce(). Probably you use the some library provide it.

answered Dec 26, 2014 at 12:43

Vladimir Bredihin

663 bronze badges

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