javascript loop through array with "for" and "ForEach"

Question

i'm trying to create a function that take an array as an argument, and return "true" only if all the items inside the array are the same .
i try to use for loop and i try to use forEeach loop
the first one work great.
the second one not!.
why?
this my first code:

function isUniform(ary) {
    var first = ary[0];
    for (i = 0; i < ary.length; i++) {
        if (first !== ary[i]) {
            return false;
        }
    }
        return true;
}
console.log(isUniform([1, 2, 1]));

this my second one:

function isUniform(ary) {
    var first = ary[0];
    ary.forEach(function(element) {
        if (first !== element) {
            return false;
        }
    });
    return true;

}
 console.log(isUniform([1, 2, 1]));

How about logging the values in the forEach loop and checking what they are? — Chris Satchell
– Chris Satchell, Commented Jun 24, 2017 at 23:45
@ChrisSatchell that's irrelevant ... the issue is there is nowhere to return to in forEach and return of an inner closure doesn't return to the outer one — charlietfl
– charlietfl, Commented Jun 24, 2017 at 23:46

Ray Toal · Accepted Answer · 2017-06-25 04:20:11Z

7

The "inner" return in your second example is a return from the function passed to forEach. It does not cause an immediate return from the isUniform function. Instead, this little anonymous function you passed to forEach runs happily to completion for every single element of your array.

After it finishes running through all of the elements, your isUniform function then simply returns true.

edited Jun 25, 2017 at 4:20

answered Jun 24, 2017 at 23:43

Ray Toal

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Comments

Rick · Accepted Answer · 2017-06-26 05:18:37Z

1

By returning true you are returning true to the forEach function callback while you're for loop returns true to the function. Your forEach requires a reference object when executing a callback if you want to use your current approach. That is how it is different from your for loop.

function isUniform(ary) {
  var first = ary[0];
  val = true;
  ary.forEach(element => {
    if (first !== element) {
      val = false;
    }
  });
  return val;
}
console.log(isUniform([1, 1, 1]));

console.log(isUniform([1, 2, 1]));

There is already an every function you can take advantage of.

arr1 = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10];

arr2 = [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1];

console.log(arr1.every((a, _, aa) => aa[0] === a));
console.log(arr2.every((a, _, aa) => aa[0] === a));

edited Jun 26, 2017 at 5:18

answered Jun 24, 2017 at 23:44

Rick

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6 Comments

user47589 Over a year ago

While true this doesn't answer the question. Why is forEach different?

tevemadar Over a year ago

With that val you should really stick to the lexical closure, just as you do with first. Bringing this into the picture pollutes the context and also makes the code confusing to read.

tevemadar Over a year ago

@Amy because of design/definition. forEach is not a conditional construct, and it does not care/check/forward/store/accumulate/etc. whatever the called function returns. If one really wants to break from a forEach loop, throwing an exception can do that, but it is not a nice solution.

Ray Toal Over a year ago

Actually Amy wasn't really asking a question here, but rather pointing out, via a rhetorical device, that the answer here simply gave a forEach-based solution to a piece of code while the OP specifically asked "WHY does the forEach code fail while the for-loop code succeeds?" Cheers.

user47589 Over a year ago

@tevemadar I know this. I was prompting him to actually answer the question.

|

Prakash Sundar · Accepted Answer · 2017-06-25 00:02:22Z

0

You can't break from a forEach.

Try this one. You can do it with the flag.

var falg = false;

function isUniform(ary) {

var first = ary[0];

ary.forEach(function(element) {

    if (first !== element) {

        falg = true;

        return false;

    }

});

if(falg)

    return false

return true;

}

console.log(isUniform([1, 2, 1]));

answered Jun 25, 2017 at 0:02

Prakash Sundar

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Comments

tevemadar · Accepted Answer · 2017-06-26 12:43:34Z

The how part has been addressed in @Arrow's post already, if you work with forEach, it does not care about the return values, you have to accumulate the result yourself (also, there is no 'normal' way of breaking out from the loop, though you could throw an exception if you really want to).

For the 'why' part: because of definition. forEach does not care about return values and runs the passed function for all elements, unconditionally.

A simplified implementation of forEach is the following:

Array.prototype.myForEach=function(callback){ // or callback,diz
  for(var i=0;i<this.length;t++)
    callback(this[i],i,this);                 // or callback.call(diz,this[i],i,this);
}

The result of callback is completely ignored. (Commented parts add support for the optional context-argument)

A 'more official' polyfill does some extra checks, but it is not complicated either: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/forEach#Polyfill

Adrià Cabezuelo · Accepted Answer · 2017-06-24 23:48:15Z

-1

can you try this, change the var element for x, and use lambdas

    function isUniform(ary) {
     var first = ary[0];
     ary.forEach((x) => {
         if(first !== x){
             return false;
         }
     });
     return true;
    }

answered Jun 24, 2017 at 23:48

Adrià Cabezuelo

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