What is happening in this loose equality comparison of 2 empty arrays

Question

I am struggling with understanding how this snippet works on a basic level

if([] == ![]){
    console.log("this evaluates to true");
}

Please help me understand where I got it wrong. My thinking:

1. First there is operator precedence so ! evaluates before ==.
2. Next ToPrimitive is called and [] converts to empty string.
3. ! operator notices that it needs to convert "" into boolean so it takes that value and makes it into false then negates into true.
4. == prefers to compare numbers so in my thinking true makes 1 and [] is converted into "" and then 0

Why does it work then? Where did I get it wrong?

Answer 1

Why does it work then?

TLDR:

[] == ![]
        //toBoolean [1]
[] == !true
[] == false
//loose equality round one [2]
//toPrimitive([]); toNumber(false) [3]
"" == 0
//loose equality round two
//toNumber("") [4]
0 === 0
true

Some explanations:

1)

First there is operator precedence so ! evaluates before ==

Negating something calls the internal toBoolean method onto that "something" first. In this case this is an Object (as Arrays are Objects) and for that it always returns true which is then negated.

2)

Now it's up to loose equalities special behaviour ( see Taurus answer for more info) :

If A is an Object ( Arrays are Objects ) and B is a Boolean it will do:

ToPrimitive(A) == ToNumber(B)

3)

  toPrimitive([])

ToPrimitive(A) attempts to convert its Object argument to a primitive value, by attempting to invoke varying sequences of A.toString and A.valueOf methods on A.

Converting an Array to its primitive is done by calling toString ( as they don't have a valueOf method) which is basically join(",").

toNumber(false)

The result is 1 if the argument is true. The result is +0 if the argument is false. Reference

So false is converted to +0

4)

toNumber("")

A StringNumericLiteral that is empty or contains only white space is converted to +0.

So finally "" is converted to +0

Where did I get it wrong?

At step 1. Negating something does not call toPrimitive but toBoolean ...

Answer 2

The accepted answer is not correct (it is now, though), see this example:

if([5] == true) {
console.log("hello"); 
}

If everything is indeed processed as the accepted answer states, then [5] == true should have evaluated to true as the array [5] will be converted to its string counterpart ("5"), and the string "5" is truthy (Boolean("5") === true is true), so true == true must be true.

But this is clearly not the case because the conditional does not evaluate to true.

So, what's actually happening is:

1. ![] will convert its operand to a boolean and then flip that boolean value, every object is truthy, so ![] is going to evaluate to false.

At this point, the comparison becomes [] == false


2. What gets into play then is these 2 rules, clearly stated in #6 in the specs for the Abstract Equality Comparison algorithm:

1. If Type(x) is boolean, return the result of the comparison ToNumber(x) == y.
2. If Type(y) is boolean, return the result of the comparison x == ToNumber(y)

At this point, the comparison becomes [] == 0.


3. Then, it is this rule:

If Type(x) is Object and Type(y) is either String or Number, return the result of the comparison ToPrimitive(x) == y.

As @Jonas_W stated, an array's ToPrimitive will call its toString, which will return a comma-separated list of its contents (I am oversimplifying).

At this point, the comparison becomes "" == 0.


4. And finally (well, almost), this rule:

If Type(x) is String and Type(y) is Number, return the result of the comparison ToNumber(x) == y.

An empty string converted to a number is 0 (Number("") == 0 is true).

At this point, the comparison becomes 0 == 0.


5. At last, this rule will apply:

If Type(x) is the same as Type(y), then
.........
  If Type(x) is Number, then
.........
    If x is the same Number value as y, return true.

---

And, this is why the comparison evaluates to true. You can also apply these rules to my first example to see why it doesn't evaluate to true.

All the rules I quoted above are clearly stated here in the specs.

Answer 3

First, realize that you are comparing two different types of values. When you use ![] it results in false. You have code (at core it results in this):

if([] == false)

Now the second part: since both values are of different type and you are using "==", javascript converts both value type into string (to make sure that they have same type). It starts from left. On the left we have []. So javascript applies the toString function on [], which results in false. Try console.log([].toString())

You should get false on the left as string value. On the right we have boolean value false, and javascript does the same thing with it. It use the toString function on false as well, which results in string value false. Try console.log(false.toString())

This involves two core concepts: how "==" works and associativity of "==" operator - whether it starts from left or right.

Associativity refers to what operator function gets called in: left-to-right or right-to-left. Operators like == or ! or + are functions that use prefix notation! The above if statement will result in false if you use "===". I hope that helps.