How does (A == B == C) comparison work in JavaScript?

I was expecting the following comparison to give an error:

var A = B = 0;
if(A == B == 0)
    console.log(true);
else
    console.log(false);

but strangely it returns false.

Even more strangely,

console.log((A == B == 1)); 

returns true.

How does this "ternary" kind of comparison work?

First, we need to understand that a == comparison between a number and a boolean value will result in internal type conversion of Boolean value to a number (true becomes 1 and false becomes 0)

The expression you have shown is evaluated from left to right. So, first

A == B

is evaluated and the result is true and you are comparing true with 0. Since true becomes 1 during comparison, 1 == 0 evaluates to false. But when you say

console.log((A == B == 1));

A == B is true, which when compared with number, becomes 1 and you are comparing that with 1 again. That is why it prints true.

Assignment operators like = are right-associative: when there is a series of these operators that have the same precedence, they are processed right-to-left, so A = B = 0 is processed as A = (B = 0) (B = 0 returns 0, so both A and B end up as 0).

Equality operators like == are left-associative: same-precedence operators are processed left-to-right. A == B == 0 is processed as (A == B) == 0, and since A == B is true (1), it becomes 1 == 0, which is false (0).

Likewise, A == B == 1 is processed as (A == B) == 1, which becomes 1 == 1, which is true (1).

Source and more info: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Operator_Precedence

First, A == B returns true, which is then compared to 0, true == 0 which returns false, or true == 1 which returns true.