JS object null checking - weird JS problem [duplicate]

Question

Imagine this simple scenario. I have variable that can be plain JS object with one property, ID, that is a number or obj variable can be null. I have simple test() function that checks if the variable is not null and that it must have valid id property.

var obj = { id: 111 };
function test() {
    return (obj && obj.id);
}

I am expecting that this function will always return boolean but in fact it returns undefined if the obj is undefined or value of obj.id if object exists like in case above. Why this function return 111 instead of true.

I am going to rip off hair of my head ... Please illuminate my mind :)

What this code says is "if obj exists, return obj.id". You could do obj && obj.id !== null — Toby
– Toby, Commented Dec 3, 2018 at 15:52
What you want is to add !! which will force it to convert to boolean. return !!(obj && obj.id) — Oram
– Oram, Commented Dec 3, 2018 at 15:52

Nurbol Alpysbayev · Accepted Answer · 2018-12-03 16:00:46Z

4

It's a common misconception. In JS (unlike in e.g. PHP) an expression like x && y does this:

execute the expression x
if the expression x returned true, then execute the expression y as well and return it (y). Otherwise return x (which would be falsy in this case e.g. 0, '', false, null, undefined).

In other words it works more like a ternary expression x ? y : z.

If you want a boolean, then use !!(x && y).

edited Dec 3, 2018 at 16:00

answered Dec 3, 2018 at 15:55

Nurbol Alpysbayev

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7 Comments

Yeldar Kurmangaliyev Over a year ago

Good explanation, but what is z in the ternary expression?

Nurbol Alpysbayev Over a year ago

@YeldarKurmangaliyev Seriously? ) I've added the ternary expression as an example to demonstrate that binary expression x && y is a conditional logical expression just like a ternary expression. z is an alternative case of ternary logical expression.

Tyler Roper Over a year ago

I think @YeldarKurmangaliyev was asking a simple (and good) question. Perhaps a more apt example would be x ? y : x, as the && operator will return x if it is falsy.

Nurbol Alpysbayev Over a year ago

@TylerRoper The "Seriously" was just my reaction after I saw his rating and impressive JS experience. He is also is the first person whom I saw here from my country BTW

Tyler Roper Over a year ago

@NurbolAlpysbayev Understood :) Simply a hiccup in communication!

|

Community · Accepted Answer · 2020-06-20 09:12:55Z

0

When `obj` is defined, why does `if (obj && obj.id)` return `111`?

The logical AND (&&) operator returns expr1 if it can be converted to false; otherwise, returns expr2.

^{^{MDN: Logical Operators - Description (slightly paraphrased)}}

expr1 (obj) cannot be converted to false, therefore it returns expr2 (111).

Why does it not return `true`?

Logical operators are typically used with Boolean (logical) values. When they are, they return a Boolean value. However, the && and || actually return the value of one of the specified operands, so if these operators are used with non-Boolean values, they may return a non-Boolean value.

^{^{MDN: Logical Operators}}

Because you are using the logical operator with non-Boolean values, the result will be non-Boolean as well.

edited Jun 20, 2020 at 9:12

CommunityBot

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answered Dec 3, 2018 at 15:53

Tyler Roper

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When `obj` is defined, why does `if (obj && obj.id)` return `111`?

Why does it not return `true`?

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When obj is defined, why does if (obj && obj.id) return 111?

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When `obj` is defined, why does `if (obj && obj.id)` return `111`?

Why does it not return `true`?