TS2367: Support loose equivalence of 'number' and 'string' #30655

TypeScript Version: 3.4.0-dev.201xxxxx

Search Terms:

• TS2367: This condition will always return 'false' since the types 'number' and 'string' have no overlap.

Code

declare var aa: string[];
declare var bb: string[];

for (let x = 0; aa[x] && bb[x]; x++) {
	if (aa[x] !== bb[x]) {
		const c	= Number(aa[x]);
		const d	= Number(bb[x]);

		if (c == aa[x] && d == bb[x]) {
			// Do stuff if both strings contain numbers
		}
	}
}

Expected behavior:

TypeScript considers loose equivalence of string and number valid, at least when type checking JavaScript code.

Actual behavior:

.hidden/stuff.ts:9:7 - error TS2367: This condition will always return 'false' since
                       the types 'number' and 'string' have no overlap.

9       if (c == aa[x] && d == bb[x]) {
            ~~~~~~~~~~    ~~~~~~~~~~

Related Issues:

• "This condition will always return 'false'" but returns true at runtime #26592
• (Trade-offs in Control Flow Analysis #9998)

• TS2367: This condition will always return 'false' #29822
• This condition will always return 'false' since the types 'true' and 'false' have no overlap #29155
• TS2367: This condition will always return 'false' since the types 'Constructor<T>' and 'typeof Child' have no overlap. #27910
• This condition will always return 'false' since the types have no overlap. Again. #25830

Additional info:

This is a snippet of legacy code that I’m not allowed to change.