Short circuit Array.forEach like calling break

Solution 1:

There's no built-in ability to break in forEach. To interrupt execution you would have to throw an exception of some sort. eg.

var BreakException = {};

try {
  [1, 2, 3].forEach(function(el) {
    console.log(el);
    if (el === 2) throw BreakException;
  });
} catch (e) {
  if (e !== BreakException) throw e;
}

JavaScript exceptions aren't terribly pretty. A traditional for loop might be more appropriate if you really need to break inside it.

Use Array#some

Instead, use Array#some:

[1, 2, 3].some(function(el) {
  console.log(el);
  return el === 2;
});

This works because some returns true as soon as any of the callbacks, executed in array order, return true, short-circuiting the execution of the rest.

some, its inverse every (which will stop on a return false), and forEach are all ECMAScript Fifth Edition methods which will need to be added to the Array.prototype on browsers where they're missing.

Solution 2:

There is now an even better way to do this in ECMAScript2015 (aka ES6) using the new for of loop. For example, this code does not print the array elements after the number 5:

let arr = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10];
for (let el of arr) {
  console.log(el);
  if (el === 5) {
    break;
  }
}

From the docs:

Both for...in and for...of statements iterate over something. The main difference between them is in what they iterate over. The for...in statement iterates over the enumerable properties of an object, in original insertion order. The for...of statement iterates over data that iterable object defines to be iterated over.

Need the index in the iteration? You can use Array.entries():

for (const [index, el] of arr.entries()) {
  if ( index === 5 ) break;
}

Solution 3:

You can use every method:

[1,2,3].every(function(el) {
    return !(el === 1);
});

ES6

[1,2,3].every( el => el !== 1 )

for old browser support use:

if (!Array.prototype.every)
{
  Array.prototype.every = function(fun /*, thisp*/)
  {
    var len = this.length;
    if (typeof fun != "function")
      throw new TypeError();

    var thisp = arguments[1];
    for (var i = 0; i < len; i++)
    {
      if (i in this &&
          !fun.call(thisp, this[i], i, this))
        return false;
    }

    return true;
  };
}

more details here.