JavaScript ‘in’ operator for `undefined` elements in Arrays

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Arrays are nothing but normal JavaScript objects, with some specialization. Quoting from Array Objects section of ECMA 5.1 Specification

Array objects give special treatment to a certain class of property names. A property name P (in the form of a String value) is an array index if and only if ToString(ToUint32(P)) is equal to P and ToUint32(P) is not equal to 232−1.

So, array indices are nothing but properties of array objects. Now, lets see about missing elements in an Array.

Quoting from the ECMA 5.1 Standard Specification,

Array elements may be elided at the beginning, middle or end of the
element list. Whenever a comma in the element list is not preceded by
an AssignmentExpression (i.e., a comma at the beginning or after
another comma), the missing array element contributes to the length of
the Array and increases the index of subsequent elements. Elided array
elements are not defined. If an element is elided at the end of an
array, that element does not contribute to the length of the Array.

So, when you say

b = [1,,,,3];

except the elements 1 and 3, all others are treated as missing elements in the array. Now elements 1 and 3 correspond to the properties 0 and 4 (Array indices start with 0 in JavaScript).

You can check that like this

console.log(Object.getOwnPropertyNames(b));
# [ '0', '4', 'length' ]
console.log(Object.getOwnPropertyNames(a));
# [ '0', '1', '2', '3', '4', 'length' ]

Only the indices 0 and 4 are in b. You might be wondering why a has the properties from 0 to 4, though the values are undefined. Because, the elements at indices 1, 2 and 3 are defined to be undefined, where as in b, we don’t know what those values are. That is why they are not allocated to a property (index).

Now, you are checking if 1 is in b, with in operator. Quoting from in operator MDN Doc,

The in operator returns true if the specified property is in the specified object.

So, you are basically checking if 1 is one of the properties of b, which is not. That is why '1' in b returns false.

Note:

If you wanted to know if 1 is in the Array or not, you should use Array.prototype.indexOf, like this

console.log(a.indexOf(1));
# 0
console.log(b.indexOf(1));
# 0
console.log(b.indexOf(5));
# -1

Array.prototype.indexOf returns -1 if the element being searched is not there in the Array. So, you might want to do something like this

console.log(a.indexOf(1) !== -1);
# true
console.log(b.indexOf(1) !== -1);
# true
console.log(a.indexOf(5) !== -1);
# false

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