Number type narrowing from `number|undefined` within arithmetic conditions

[Knagis]

TypeScript Version: 2.1.5

JavaScript evaluates undefined > 0 (and other similar constructs) as false and this could be used for type narrowing. Unlike comparisons with null, undefined is rather consistent (https://jsfiddle.net/4u11p7u0/1/).

Code

let a: number | undefined;

function expectNumber(a: number) { }

if (a != null) {
    // works
    expectNumber(a);    
}

if (a > 0) {
    // does not work but should because "undefined > 0 === false"
    expectNumber(a);
}

Expected behavior:

the if (a > 0) branch narrows the type to number.

Actual behavior:

the type of a within the branch stays as number|undefined.

[ahejlsberg]

We're actually going in the opposite direction on this one. In the upcoming TypeScript 2.2 release, for an operand a that is possibly undefined we consider the expression a > 0 to be an error. See #13483 for more details.

Generally the results of the <, >, <=, and >= operators with null and undefined operands are surprising and merit reporting an error. For example, a > 0 and a < 0 are both false when a is undefined.

[Knagis]

Thank you for the information.