Why `null >= 0 && null <= 0` but not `null == 0`?

Question

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.

I'd like to extend the question to further improve visibility of the problem:

null >= 0; //true
null <= 0; //true
null == 0; //false
null > 0;  //false
null < 0;  //false

It just makes no sense. Like human languages, these things need be learned by heart.

JavaScript has both strict and type–converting comparisons

null >= 0; is true but (null==0)||(null>0) is false

null <= 0; is true but (null==0)||(null<0) is false

"" >= 0 is also true

For relational abstract comparisons (<= , >=), the operands are first converted to primitives, then to the same type, before comparison.

typeof null returns "object"

When type is object javascript tries to stringify the object (i.e null) the following steps are taken (ECMAScript 2015):

1. If PreferredType was not passed, let hint be "default".
2. Else if PreferredType is hint String, let hint be "string".
3. Else PreferredType is hint Number, let hint be "number".
4. Let exoticToPrim be GetMethod(input, @@toPrimitive).
5. ReturnIfAbrupt(exoticToPrim).
6. If exoticToPrim is not undefined, then
   a) Let result be Call(exoticToPrim, input, «hint»).
   b) ReturnIfAbrupt(result).
   c) If Type(result) is not Object, return result.
   d) Throw a TypeError exception.
7. If hint is "default", let hint be "number".
8. Return OrdinaryToPrimitive(input,hint).

The allowed values for hint are "default", "number", and "string". Date objects, are unique among built-in ECMAScript object in that they treat "default" as being equivalent to "string". All other built-in ECMAScript objects treat "default" as being equivalent to "number". (ECMAScript 20.3.4.45)

So I think null converts to 0.

console.log( null > 0 );  // (1) false
console.log( null == 0 ); // (2) false
console.log( null >= 0 ); // (3) true

Mathematically, that’s strange. The last result states that "null is greater than or equal to zero", so in one of the comparisons above it must be true, but they are both false.

The reason is that an equality check == and comparisons > < >= <= work differently. Comparisons convert null to a number, treating it as 0. That’s why (3) null >= 0 is true and (1) null > 0 is false.

On the other hand, the equality check == for undefined and null is defined such that, without any conversions, they equal each other and don’t equal anything else. That’s why (2) null == 0 is false.

I had the same problem !!. Currently my only solution is to separate.

var a = null;
var b = undefined;

if (a===0||a>0){ } //return false  !work!
if (b===0||b>0){ } //return false  !work!

//but 
if (a>=0){ } //return true !