3 ms·

The problem with using any where you don't care about the type is that someone can come along and make unchecked assumptions about what type it actually is. Un

The problem with using any where you don't care about the type is that someone can come along and make unchecked assumptions about what type it actually is. Unknown is spicier as it'll prevent developers from assuming it's a string or object or something.

jax_the_dog 4y ago
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.)

jax_the_dog 4y ago

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.)