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That makes some sense to me. I’m not a TypeScript dev so I apologize for the stupid question. Is there any functional difference between the two? I.e. is there

That makes some sense to me. I’m not a TypeScript dev so I apologize for the stupid question. Is there any functional difference between the two? I.e. is there a case where using Unknown instead of Any will result in some sort of “compile” time error opposed to a runtime?

thoughtspile 4y ago
Almost every case where you use the value results in an error with unknown, but not with any: let danger: any; // These all compile: danger(), 9 / danger, danger.access, danger.map(x => x \* 2) let safe: unknown; // These all explode in TS: safe(), 9 / safe, safe.access, safe.map(x => x \* 2)
madeofpalk 4y ago
Yes - trying to access a property or method on something typed as unknown will (almost?) always result in a compile time error because it's unknown whether the value has that property or not. https://www.typescriptlang.org/play?#code/GYVwdgxgLglg9mABAdwE4EMAOBhBBnOAGwFMAZOAcwEEwBPACigAsYwKAVOcigLkXToBKRAG8AUIkQR8RYgDpClegCIAsrUQB1YgCNEVTJmUAaRM1YculOVDgBVQ8VTZ0eYvUGCxAXzFjQkLAIKBg4MiTcdmAA1mBwyGCMLGyc3HzgsfFgwuKS0mAEJApKahraegZGpuYpVhQ29o7Oru6ePn4B0PBIaFi4BbKRMXEJAELEUFBOSRaplOnDWTkSiDDAiIy0mMRw6zWW3IgAvCeIynhQqBbKy5J54fKKFCrqWrr6hiZmyQfWtg7bZpuDxeSTeRDEQhuUQre4DIpPF5ld6VL77OYUUGIXzeIA https://www.typescriptlang.org/play?#code/GYVwdgxgLglg9mABAd...
- thoughtspile 4y ago
  always, because "unknown" includes null / undefined, which don't allow property access at all.
WorldMaker 4y ago
Yes. `any` really does turn off all type checking. You see that offhand mentioned in this article talking about some of the paradoxes of `any`. `unknown` is still type checked and is arguably "merciless" type-checked that to do much of anything with an `unknown` you have to check for a more specific type first or the compiler returns an error that what you are doing isn't known to be valid for `unknown`. Some of that happens "automatically" at this point in the large number of ways that types can now be narrowed implicitly in Typescript (type guards [library functions], type asserts, the `typeof` runtime operator, the `in` runtime operator [as of recently], etc), so it can feel like `unknown`/`any` are the same up to a point, that point being where the runtime type is trivially known based on if statements and library functions around your use of the type. (Fun fact: `any` predates most type narrowing and `unknown` by several major Typescript versions. So `any` beyond just being the final "escape hatch" from type checking is also something of a legacy tool.)

thoughtspile 4y ago

Almost every case where you use the value results in an error with unknown, but not with any: let danger: any; // These all compile: danger(), 9 / danger, danger.access, danger.map(x => x \* 2) let safe: unknown; // These all explode in TS: safe(), 9 / safe, safe.access, safe.map(x => x \* 2)

madeofpalk 4y ago

Yes - trying to access a property or method on something typed as unknown will (almost?) always result in a compile time error because it's unknown whether the value has that property or not. https://www.typescriptlang.org/play?#code/GYVwdgxgLglg9mABAdwE4EMAOBhBBnOAGwFMAZOAcwEEwBPACigAsYwKAVOcigLkXToBKRAG8AUIkQR8RYgDpClegCIAsrUQB1YgCNEVTJmUAaRM1YculOVDgBVQ8VTZ0eYvUGCxAXzFjQkLAIKBg4MiTcdmAA1mBwyGCMLGyc3HzgsfFgwuKS0mAEJApKahraegZGpuYpVhQ29o7Oru6ePn4B0PBIaFi4BbKRMXEJAELEUFBOSRaplOnDWTkSiDDAiIy0mMRw6zWW3IgAvCeIynhQqBbKy5J54fKKFCrqWrr6hiZmyQfWtg7bZpuDxeSTeRDEQhuUQre4DIpPF5ld6VL77OYUUGIXzeIA https://www.typescriptlang.org/play?#code/GYVwdgxgLglg9mABAd...

thoughtspile 4y ago
always, because "unknown" includes null / undefined, which don't allow property access at all.

WorldMaker 4y ago

Yes. `any` really does turn off all type checking. You see that offhand mentioned in this article talking about some of the paradoxes of `any`. `unknown` is still type checked and is arguably "merciless" type-checked that to do much of anything with an `unknown` you have to check for a more specific type first or the compiler returns an error that what you are doing isn't known to be valid for `unknown`. Some of that happens "automatically" at this point in the large number of ways that types can now be narrowed implicitly in Typescript (type guards [library functions], type asserts, the `typeof` runtime operator, the `in` runtime operator [as of recently], etc), so it can feel like `unknown`/`any` are the same up to a point, that point being where the runtime type is trivially known based on if statements and library functions around your use of the type. (Fun fact: `any` predates most type narrowing and `unknown` by several major Typescript versions. So `any` beyond just being the final "escape hatch" from type checking is also something of a legacy tool.)