How does JS type coercion work?

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"0" is a string containing the character 0, it is not the numeric value 0. The only string-type value which evaluates to false is "".

"0" is truthy.

Section 9.2 of the ECMAScript 262 specification defines how different types are converted to Boolean:

Argument Type   Result
Undefined       false
Null            false
Boolean         The result equals the input argument (no conversion).
Number          The result is false if the argument is +0, −0, or NaN; otherwise the
                result is true.
String          The result is false if the argument is the empty String (its length is
                zero); otherwise the result is true.
Object          true

This, however, is only strictly followed when comparing using ===.

When using Boolean('0') you’re converting the value '0' to Boolean (which is the same as using !!'0'). When loosely comparing '0' with false, the Boolean value is converted to a number (as defined here). false, when converted to a number, becomes 0. This means the final calculation is '0' == 0 which equates to true.

To summarise the relevant part of the linked section of the ECMAScript specification above:

  1. Let x = '0' and y = false.
  2. Check if the type of y is Boolean.
  3. If true, convert y to a number.
  4. Compare x to the numeric equivalent of y.

In our case, a JavaScript implementation of this would be:

var x = '0',                      // x = "0"
    y = false;                    // y = false

if (typeof y === "boolean") {
    y = +y;                       // y = 0
}

console.log( x == y );            // "0" == 0

-> true

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