map did not return array when has an inner loop

Question

const testTrue = sample.map((condition) => {
  return Object.values(objects).forEach(value => {
    return someFunc(condition.abc, value)
  })
});

sample is an array, why above code don't make testTrue become [ true, false ] ? I got undefined, is it because of forEach here?

This is working

const testTrue = sample.map((condition) => {
  return someFunc(condition.abc, value)
});

The value returned to the forEach callback is ignored. What were you trying to do there? Can you give an example of condition and objects and your expected output? — CertainPerformance
– CertainPerformance, Commented Jul 12, 2018 at 6:29
Can you also provide info on what will be present in sample array? — Niragh
– Niragh, Commented Jul 12, 2018 at 6:36
please post your someFunc definition and how you want the result — anonymous
– anonymous, Commented Jul 12, 2018 at 7:38

cнŝdk · Accepted Answer · 2018-07-12 06:39:59Z

3

Actually, Array#forEach() doesn't return anything, if you take a look at the Array#forEach() MDN Specifictaion you can see that its ReturnValue is undefined.

And that's why the map call isn't returning any element, in your code:

const testTrue = sample.map((condition) => {
  return Object.values(objects).forEach(value => {
    //The following will be ignored
    return someFunc(condition.abc, value)
  })
});

The call to return someFunc(condition.abc, value) is useless and it is as if you weren't writing anything.

Solution:

You can change the forEach() with an Array#map() call to return transformed elements of your array:

const testTrue = sample.map((condition) => {
  return Object.values(objects).map(value => {
    return someFunc(condition.abc, value)
  })
});

edited Jul 12, 2018 at 6:39

answered Jul 12, 2018 at 6:34

cнŝdk

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1 Comment

Alisa F. Kennedy Over a year ago

I got array within array using your solution.

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