[] == false ??

JavaScriptJunior~1 min read

Short answer

Because when comparing with ==, JavaScript automatically coerces both sides to a number, and ultimately compares 0 == 0.

Step-by-step explanation

Let's take:

javascript

[] == false

Step 1. `==` triggers type coercion

With loose comparison (==), JS tries to coerce the types so the comparison is correct. The algorithm is described in the ECMAScript standard §7.2.14 as follows:

If one of the operands is boolean, it is coerced to a number.

Step 2. Convert `false` → to a number

javascript

false → 0

Now the expression looks like this:

javascript

[] == 0

Step 3. JS sees that the left side is an object (array), and the right side is a primitive

If one of the operands is an object, it is first coerced to a primitive (via valueOf() → toString()).

For the array []:

javascript

[].toString(); // ""

So:

javascript

[] → ""  // empty string

Now we have:

javascript

"" == 0

Step 4. Comparing `"" == 0`

According to the same algorithm:

If one operand is a string and the other is a number → the string is converted to a number.

javascript

Number("") → 0

So:

javascript

0 == 0

Result:

javascript

true

Summary by steps

Step	Expression	What happened
1	`[] == false`	`false → 0`
2	`[] == 0`	`[] → ""`
3	`"" == 0`	`"" → 0`
4	`0 == 0`	`true`

Why this is dangerous

This logic makes == unpredictable. Here are more examples:

javascript

[] == ![]     // true
'' == 0       // true
'0' == 0      // true
null == undefined // true
[] == ''      // true
[0] == 0      // true
[1] == 1      // true
[2] == 2      // true

How to avoid confusion

Use strict comparison ===, which does not perform implicit type conversion.

javascript

[] === false // false
"" === 0     // false
null === undefined // false

In short

Expression	What JS does	Result
`[] == false`	`[] → "" → 0`, `false → 0`	`true`
`[] === false`	Compares without coercion	`false`

Short Answer

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