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MartinJohns avatar MartinJohns commented on June 3, 2024 2

This is explicitly out of scope for TypeScript. Check the "Viability Checklist" again, specifically "emitting based on the type of expressions", "this isn't a runtime feature" and the Design Goals, goal 9 and non-goal 5. It's also been suggested and declined numerous times already.

While libraries like Zod allow type inference, it's not possible to infer TypeScript interfaces from schemas,

That doesn't seem to be accurate. You should be able to do this: type User = z.infer<typeof UserSchema>;

from typescript.

NinjaKinshasa avatar NinjaKinshasa commented on June 3, 2024

This is explicitly out of scope for TypeScript. Check the "Viability Checklist" again, specifically "emitting based on the type of expressions", "this isn't a runtime feature" and the Design Goals, goal 9 and non-goal 5. It's also been suggested and declined numerous times already.

Ok I didn't know about the design goals page , thank you for pointing it out. This feature would not keep typescript "fully erasable". I close this issue.

While libraries like Zod allow type inference, it's not possible to infer TypeScript interfaces from schemas,

That doesn't seem to be accurate. You should be able to do this: type User = z.infer;

You are correct I switched the words, I edited my first message to fix it.

from typescript.

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