Why does foreach skip one element from array

Question

I am trying to iterate an array with forEach and, based on a condition, I do something.

For this question, I simplified the condition in order to try to understand what's going on.

I expect array b = [] after the operation, but it is not, as it operates only on half on the elements. Why does this happen?

(This is not about removing everything from b, just trying to understand why it jumps the even indexes).

var a = [1, 2, 3, 4, 5, 6];
var b = a.slice(0);

console.log('before b = ', b); // b = [1, 2, 3, 4, 5, 6]

a.forEach(function (e) {
    if (e > 0) {
        b.splice(a.indexOf(e), 1);
    }
});


console.log('after b = ', b); // b = [2, 4, 6]

// but I expect b = []

Giannis Mp · Accepted Answer · 2018-03-04 16:37:03Z

3

It does not. It goes through each and every item. The thing is what you do with array b. First you remove the index 0 from it which is 1. So now b = [2,3,4,5,6]. Then index 1 which is 3 so b = [2,4,5,6]. Then index 2 which is 5 so b = [2,4,6]. The next indexes don't exist so in the end b = [2,4,6].

To have the expected outcome use b.indexOf(e) in your splice call.

var a = [1, 2, 3, 4, 5, 6];
var b = a.slice(0);

console.log('before b = ', b); // b = [1, 2, 3, 4, 5, 6]

a.forEach(function (e) {
    if (e > 0) {
        b.splice(b.indexOf(e), 1);
    }
});


console.log('after b = ', b); // b = [2, 4, 6]

// but I expect b = []

answered Mar 4, 2018 at 16:37

Giannis Mp

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3 Comments

DonJoe Over a year ago

Thanks. I understand now.

DonJoe Over a year ago

I feel so dumb now. The thing is that, initially, there was only a array and I was slicing it directly, but still getting only half of the expected results. Later I added b in order to understand what was happening. I think there was another problem inside the function which made it skip elements.

Giannis Mp Over a year ago

You are welcome, you have to be careful when you splice an array within the forEach

Nina Scholz · Accepted Answer · 2018-03-04 16:35:19Z

2

You could take the index directly for splicing and make an log from the value an the array. You se, that the array becomes smaller than the index for splicing.

var a = [1, 2, 3, 4, 5, 6];
var b = a.slice(0);

console.log(b.join(' '));

a.forEach(function (e, i) {
    if (e > 0) {
        b.splice(a.indexOf(e), 1);
        console.log(i, ': ', b.join(' '));
    }
});

console.log(b); // b = [2, 4, 6]

To empty b, you need to look for the index of the array b, not a and splice it.

var a = [1, 2, 3, 4, 5, 6];
var b = a.slice(0);

a.forEach(function (e, i) {
    if (e > 0) {
        b.splice(b.indexOf(e), 1);
        //       ^
    }
});

console.log(b);

answered Mar 4, 2018 at 16:35

Nina Scholz

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Comments

Alex · Accepted Answer · 2018-03-04 16:39:48Z

1

You're splicing based on the index of the element from a, however b's indices are getting updated when you splice in it. You should try to splice from b based on the index of the element IN b.

answered Mar 4, 2018 at 16:39

Alex

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