Can you please elaborate on the concept of type coercion in JavaScript?

I've come across information stating that when comparing an object with a number, type-coercion occurs. ToPrimitive is called on the object which then invokes valueOf and, if necessary, toString. However, I'm struggling to understand how this process actually works in the code snippet below:

[ null ] == 0             // true
[ undefined ] == 0        // true
[ ] == 0                  // true
[ "-0" ] == 0             // true

[ false ] == 0            // false
[ true ] == 0             // false

Why is [ false ] == 0 evaluated as false but [ null ], [ undefined ] are evaluated as true?

Answer №1

When using Number([false]), it results in NaN because it converts to Number('false'). However, '' (empty or only white-space), null, and undefined initialize a Number as 0. The Number wrapper returns a primitive value.

All primitive values have object equivalents except for null and undefined, which wrap around the primitive values.

As Touffy pointed out in the comments, the array must be resolved by using toString on the value before calling the wrapper. Once the string value is determined, the wrapper handles the parsing.

Number([0])         // [0].toString()         => Number("0")     => 0
Number([undefined]) // [undefined].toString() => Number("")      => 0
Number([null])      // [null].toString()      => Number("")      => 0 
Number([false])     // [false].toString()     => Number("false") => NaN

Both null and undefined are valid arguments for creating a Number primitive because they represent the absence of a value. Blank or empty strings are also considered empty values. Strings like 'false' and boolean values like false are non-empty and not numeric, resulting in NaN.

Tip: This differs slightly from how integer parsing functions.

parseInt('0')     // 0
parseInt('')      // NaN
parseInt('false') // NaN

You can explore

string_number_conversions_internal.h
in the Chromium Code Search to understand how parsing works for strings.

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