Question

I had to write a routine that increments the value of a variable by 1 if its type is number and assigns 0 to the variable if not, where the variable is initially null or undefined.

The first implementation was v >= 0 ? v += 1 : v = 0 because I thought anything not a number would make an arithmetic expression false, but it was wrong since null >= 0 is evaluated to true. Then I learned null behaves like 0 and the following expressions are all evaluated to true.

• null >= 0 && null <= 0
• !(null < 0 || null > 0)
• null + 1 === 1
• 1 / null === Infinity
• Math.pow(42, null) === 1

Of course, null is not 0. null == 0 is evaluated to false. This makes the seemingly tautological expression (v >= 0 && v <= 0) === (v == 0) false.

Why is null like 0, although it is not actually 0?

Answer 1

Your real question seem to be:

Why:

null >= 0; // true

But:

null == 0; // false

What really happens is that the Greater-than-or-equal Operator (>=), performs type coercion (ToPrimitive), with a hint type of Number, actually all the relational operators have this behavior.

null is treated in a special way by the Equals Operator (==). In a brief, it only coerces to undefined:

null == null; // true
null == undefined; // true

Value such as false, '', '0', and [] are subject to numeric type coercion, all of them coerce to zero.

You can see the inner details of this process in the The Abstract Equality Comparison Algorithm and The Abstract Relational Comparison Algorithm.

In Summary:

• Relational Comparison: if both values are not type String, ToNumber is called on both. This is the same as adding a + in front, which for null coerces to 0.

• Equality Comparison: only calls ToNumber on Strings, Numbers, and Booleans.