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Aren't you implicitly assuming that TS's number === JS's Number and TS's string === JS's String? You could say == is extended for TS's types(number/string), but
by Sacho 8y ago
Aren't you implicitly assuming that TS's number === JS's Number and TS's string === JS's String? You could say == is extended for TS's types(number/string), but the JS == still exists: x as any == y as any;
- kbp 8y agoI can see what you're saying, but I think that argument would be a lot stronger if == compiled to ===. If you want an operator with ==='s semantics, it's there; providing == and claiming it behaves like === just seems like an opportunity for annoying surprises if something slips through the cracks. Since == doesn't compile to ===, I wouldn't use it unless I understood its semantics and wanted them.