[javascript] Is there a mechanism to loop x times in ES6 (ECMAScript 6) without mutable variables?

The typical way to loop x times in JavaScript is:

for (var i = 0; i < x; i++)
  doStuff(i);

But I don't want to use the ++ operator or have any mutable variables at all. So is there a way, in ES6, to loop x times another way? I love Ruby's mechanism:

x.times do |i|
  do_stuff(i)
end

Anything similar in JavaScript/ES6? I could kind of cheat and make my own generator:

function* times(x) {
  for (var i = 0; i < x; i++)
    yield i;
}

for (var i of times(5)) {
  console.log(i);
}

Of course I'm still using i++. At least it's out of sight :), but I'm hoping there's a better mechanism in ES6.

Here is another good alternative:

Array.from({ length: 3}).map(...);

Preferably, as @Dave Morse pointed out in the comments, you can also get rid of the map call, by using the second parameter of the Array.from function like so:

Array.from({ length: 3 }, () => (...))

Advantages of this solution

• Simplest to read / use (imo)
• Return value can be used as a sum, or just ignored
• Plain es6 version, also link to TypeScript version of the code

Disadvantages - Mutation. Being internal only I don't care, maybe some others will not either.

Examples and Code

times(5, 3)                       // 15    (3+3+3+3+3)

times(5, (i) => Math.pow(2,i) )   // 31    (1+2+4+8+16)

times(5, '<br/>')                 // <br/><br/><br/><br/><br/>

times(3, (i, count) => {          // name[0], name[1], name[2]
    let n = 'name[' + i + ']'
    if (i < count-1)
        n += ', '
    return n
})

function times(count, callbackOrScalar) {
    let type = typeof callbackOrScalar
    let sum
    if (type === 'number') sum = 0
    else if (type === 'string') sum = ''

    for (let j = 0; j < count; j++) {
        if (type === 'function') {
            const callback = callbackOrScalar
            const result = callback(j, count)
            if (typeof result === 'number' || typeof result === 'string')
                sum = sum === undefined ? result : sum + result
        }
        else if (type === 'number' || type === 'string') {
            const scalar = callbackOrScalar
            sum = sum === undefined ? scalar : sum + scalar
        }
    }
    return sum
}

TypeScipt version
https://codepen.io/whitneyland/pen/aVjaaE?editors=0011

const times = 4;
new Array(times).fill().map(() => console.log('test'));

This snippet will console.log test 4 times.

Answer: 09 December 2015

Personally, I found the accepted answer both concise (good) and terse (bad). Appreciate this statement might be subjective, so please read this answer and see if you agree or disagree

The example given in the question was something like Ruby's:

x.times do |i|
  do_stuff(i)
end

Expressing this in JS using below would permit:

times(x)(doStuff(i));

Here is the code:

let times = (n) => {
  return (f) => {
    Array(n).fill().map((_, i) => f(i));
  };
};

That's it!

Simple example usage:

let cheer = () => console.log('Hip hip hooray!');

times(3)(cheer);

//Hip hip hooray!
//Hip hip hooray!
//Hip hip hooray!

Alternatively, following the examples of the accepted answer:

let doStuff = (i) => console.log(i, ' hi'),
  once = times(1),
  twice = times(2),
  thrice = times(3);

once(doStuff);
//0 ' hi'

twice(doStuff);
//0 ' hi'
//1 ' hi'

thrice(doStuff);
//0 ' hi'
//1 ' hi'
//2 ' hi'

Side note - Defining a range function

A similar / related question, that uses fundamentally very similar code constructs, might be is there a convenient Range function in (core) JavaScript, something similar to underscore's range function.

Create an array with n numbers, starting from x

Underscore

_.range(x, x + n)

ES2015

Couple of alternatives:

Array(n).fill().map((_, i) => x + i)

Array.from(Array(n), (_, i) => x + i)

Demo using n = 10, x = 1:

> Array(10).fill().map((_, i) => i + 1)
// [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 ]

> Array.from(Array(10), (_, i) => i + 1)
// [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 ]

In a quick test I ran, with each of the above running a million times each using our solution and doStuff function, the former approach (Array(n).fill()) proved slightly faster.

addressing the functional aspect:

function times(n, f) {
    var _f = function (f) {
        var i;
        for (i = 0; i < n; i++) {
            f(i);
        }
    };
    return typeof f === 'function' && _f(f) || _f;
}
times(6)(function (v) {
    console.log('in parts: ' + v);
});
times(6, function (v) {
    console.log('complete: ' + v);
});

I think it is pretty simple:

[...Array(3).keys()]

or

Array(3).fill()

I wrapped @Tieme s answer with a helper function.

In TypeScript:

export const mapN = <T = any[]>(count: number, fn: (...args: any[]) => T): T[] => [...Array(count)].map((_, i) => fn())

Now you can run:

const arr: string[] = mapN(3, () => 'something')
// returns ['something', 'something', 'something']

I think the best solution is to use let:

for (let i=0; i<100; i++) …

That will create a new (mutable) i variable for each body evaluation and assures that the i is only changed in the increment expression in that loop syntax, not from anywhere else.

I could kind of cheat and make my own generator. At least i++ is out of sight :)

That should be enough, imo. Even in pure languages, all operations (or at least, their interpreters) are built from primitives that use mutation. As long as it is properly scoped, I cannot see what is wrong with that.

You should be fine with

function* times(n) {
  for (let i = 0; i < n; i++)
    yield i;
}
for (const i of times(5)) {
  console.log(i);
}

But I don't want to use the ++ operator or have any mutable variables at all.

Then your only choice is to use recursion. You can define that generator function without a mutable i as well:

function* range(i, n) {
  if (i >= n) return;
  yield i;
  return yield* range(i+1, n);
}
times = (n) => range(0, n);

But that seems overkill to me and might have performance problems (as tail call elimination is not available for return yield*).

Generators? Recursion? Why so much hatin' on mutatin'? ;-)

If it is acceptable as long as we "hide" it, then just accept the use of a unary operator and we can keep things simple:

Number.prototype.times = function(f) { let n=0 ; while(this.valueOf() > n) f(n++) }

Just like in ruby:

> (3).times(console.log)
0
1
2

for (let i of Array(100).keys()) {
    console.log(i)
}

Array(100).fill().map((_,i)=> console.log(i) );

This version satisifies the OP's requirement for immutability. Also consider using reduce instead of map depending on your use case.

This is also an option if you don't mind a little mutation in your prototype.

Number.prototype.times = function(f) {
   return Array(this.valueOf()).fill().map((_,i)=>f(i));
};

Now we can do this

((3).times(i=>console.log(i)));

+1 to arcseldon for the .fill suggestion.

The simplest way I can think of for creating list/array within range

Array.from(Array(max-min+1), (_, index) => index+min)

Using the ES2015 Spread operator:

[...Array(n)].map()

const res = [...Array(10)].map((_, i) => {
  return i * 10;
});

// as a one liner
const res = [...Array(10)].map((_, i) => i * 10);

Or if you don't need the result:

[...Array(10)].forEach((_, i) => {
  console.log(i);
});

// as a one liner
[...Array(10)].forEach((_, i) => console.log(i));

Or using the ES2015 Array.from operator:

Array.from(...)

const res = Array.from(Array(10)).map((_, i) => {
  return i * 10;
});

// as a one liner
const res = Array.from(Array(10)).map((_, i) => i * 10);

Note that if you just need a string repeated you can use String.prototype.repeat.

console.log("0".repeat(10))
// 0000000000

If you're willing to use a library, there's also lodash _.times or underscore _.times:

_.times(x, i => {
   return doStuff(i)
})

Note this returns an array of the results, so it's really more like this ruby:

x.times.map { |i|
  doStuff(i)
}

I am just going to put this here. If you are looking for a compact function without using Arrays and you have no issue with mutability/immutability :

var g =x=>{/*your code goes here*/x-1>0?g(x-1):null};
 

Not something I would teach (or ever use in my code), but here's a codegolf-worthy solution without mutating a variable, no need for ES6:

Array.apply(null, {length: 10}).forEach(function(_, i){
    doStuff(i);
})

More of an interesting proof-of-concept thing than a useful answer, really.

I made this:

function repeat(func, times) {
    for (var i=0; i<times; i++) {
        func(i);
    }
}

Usage:

repeat(function(i) {
    console.log("Hello, World! - "+i);
}, 5)

/*
Returns:
Hello, World! - 0
Hello, World! - 1
Hello, World! - 2
Hello, World! - 3
Hello, World! - 4
*/

The i variable returns the amount of times it has looped - useful if you need to preload an x amount of images.

I am late to the party, but since this question turns up often in search results, I would just like to add a solution that I consider to be the best in terms of readability while not being long (which is ideal for any codebase IMO). It mutates, but I'd make that tradeoff for KISS principles.

let times = 5
while( times-- )
    console.log(times)
// logs 4, 3, 2, 1, 0

Afaik, there is no mechanism in ES6 similar to Ruby's times method. But you can avoid mutation by using recursion:

let times = (i, cb, l = i) => {
  if (i === 0) return;

  cb(l - i);
  times(i - 1, cb, l);
}

times(5, i => doStuff(i));

Demo: http://jsbin.com/koyecovano/1/edit?js,console

In the functional paradigm repeat is usually an infinite recursive function. To use it we need either lazy evaluation or continuation passing style.

Lazy evaluated function repetition

const repeat = f => x => [x, () => repeat(f) (f(x))];_x000D_
const take = n => ([x, f]) => n === 0 ? x : take(n - 1) (f());_x000D_
_x000D_
console.log(_x000D_
  take(8) (repeat(x => x * 2) (1)) // 256_x000D_
);

I use a thunk (a function without arguments) to achieve lazy evaluation in Javascript.

Function repetition with continuation passing style

const repeat = f => x => [x, k => k(repeat(f) (f(x)))];_x000D_
const take = n => ([x, k]) => n === 0 ? x : k(take(n - 1));_x000D_
_x000D_
console.log(_x000D_
  take(8) (repeat(x => x * 2) (1)) // 256_x000D_
);

CPS is a little scary at first. However, it always follows the same pattern: The last argument is the continuation (a function), which invokes its own body: k => k(...). Please note that CPS turns the application inside out, i.e. take(8) (repeat...) becomes k(take(8)) (...) where k is the partially applied repeat.

Conclusion

By separating the repetition (repeat) from the termination condition (take) we gain flexibility - separation of concerns up to its bitter end :D