RegExp.test() fails to build in strict mode with undefined or null input

[styfle]

TypeScript Version: 2.7.0-dev.20171206

Code

function getDevice() {
    if (window && window.screen) {
        const w = window.screen.width;
        if (w > 1000) {
            return 'desktop';
        } else if (w > 800) {
            return 'tablet';
        } else {
            return 'mobile';
        }
    }
    return undefined;
}

const device = getDevice();
const isMobile = /mobile|tablet/.test(device);

Expected behavior:

Should compile in strict mode.

Actual behavior:

Does not compile in strict mode.

Type 'undefined' is not assignable to type 'string'.

Workaround

The workaround is to check if undefined and return a string before calling regexp.test(). For example:

const device = getDevice() || '';
const isMobile = /mobile|tablet/.test(device);

Solution

I believe the type definition for RegExp.test() should be the following

test(string: string | undefined | null): boolean;

The reason being, that test() seem to return false in the case of null or undefined.
MDN does not mention that it throws.

[ghost]

Seems like a duplicate of #4002 -- the function is intended to be used with strings and will convert any other input to a string. It won't throw an error but probably isn't doing what you intended. (/undefined/.test(undefined) is true!)

[styfle]

@andy-ms That is very interesting. I didn't consider the case of /undefined/.test(undefined) because my example is clearly not testing against /undefined/ so there is no unintended bug.

I'll leave the reader with this real world example that is deceiving at first glance:

const reg = /(def|func)/;
console.log(reg.test('func')); // true, expected
console.log(reg.test('def')); // true, expected
console.log(reg.test('nope')); // false, expected
console.log(reg.test(undefined)); // true!, WHAAT??

I suppose TypeScript is right to complain in order to catch these odd corner cases.