How JavaScript (A == B == C) Comparison Works

comparisonjavascripttruthiness

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?

Best Answer

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.

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