Why does the array reads value not present?

Question

Why is the console displaying arrayNew? I'm hoping a condition where if the array includes the elements 'a' and 'b' only then the console displays arrayNew.Can someone help me there?

var arrayNew = ['a'];
if(arrayNew.includes('b' && 'a'))
  console.log(arrayNew);

Have you tried to check what `'b' && 'd'` yields? None of the conditions check whether two strings are part of that array, but whether the **evaluated** value of AND is part of that array — Nico Haase, Jun 29 '21 at 10:25
Javascript isn't English. Your condition does not literally mean "arrayNew includes b and d"… — deceze, Jun 29 '21 at 10:27

score 2 · Answer 1 · answered Jun 29 '21 at 10:26

&& evaluates as the left-hand value if that value is falsy and the right-hand value otherwise.

Since 'b' is a true value, 'b' && 'd' evaluates as 'd'. The array doesn't include that so arrayNew.includes('d') is false.

Since 'd' is a true value, 'd' && 'b' evaluates as 'b'. The array does include that so arrayNew.includes('b') is true.

score 1 · Answer 2 · edited Jun 29 '21 at 10:42

1

Because 'b' && 'd' === 'd', the array doesn't include 'd', so the first if is false.

The second if is true because 'd' && 'b' === 'b' and 'b' is in the array.

If you want to know if it includes both values you must do

if (arr.includes('a') && arr.includes('b'))

edited Jun 29 '21 at 10:42

deceze

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answered Jun 29 '21 at 10:30

Jazz

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Why does the array reads value not present?

2 Answers2