variables type unknown not accessible typescript

Question

What if I use a first-class function that is stored in a variable which is an unknown type. Trying to do so getting the following compiler error.

error TS2571: Object is of type 'unknown'.

    let somethingany : unknown; 
    let somethingunknown : unknown;
    const loadUnkown: unknown  = (params: unknown) : unknown=>{  <--- exactly her is the mistake
        if(typeof params == 'string'){
            return params.trim();
        }
    }
    
    
    somethingany = loadUnkown('werwq'); <----//eror here
    console.log(somethingany.trim());
    
    somethingunknown = loadUnkown('werwer');  <----//eror here
    if(typeof somethingunknown == 'string'){
        console.log(somethingunknown.trim());
    }

Though any is working precisely.

let somethingany : any;
let somethingunknown : any;
const loadString: any = (params: any): any =>{
    return params;
}

somethingany = loadString('asdasdsad');
console.log(somethingany.trim());

somethingunknown = loadString('asdasdsad');
if(typeof somethingunknown == 'string'){
    console.log(somethingunknown.trim());
}

Need to know why const loadUnkown: unknown = ()=>{} is a problem

Here is an unswer: https://stackoverflow.com/questions/51439843/unknown-vs-any — Oleksandr Sakun, Nov 27 '20 at 12:57
no, not helping not looking for the difference between any vs unknown. Need to know why const loadUnkown: unknown is a problem — ANIK ISLAM SHOJIB, Nov 27 '20 at 13:01
Because unknown is not callable. try `if(typeof loadUnkown === "function") { loadUnknown("string");}` — kevinSpaceyIsKeyserSöze, Nov 27 '20 at 13:21
You already have the answer linked but here it is again :) ['unknown' vs. 'any'](https://stackoverflow.com/questions/51439843/unknown-vs-any) — , Nov 27 '20 at 13:38
It's not the same question, but the answer is the same. Thus, it pretty much is the same question — , Nov 27 '20 at 14:11

kevinSpaceyIsKeyserSöze · Accepted Answer · 2020-11-27T14:14:06.050

As requested here is my answer. Your question does not make a lot of sense as the the variable foo already has a type in your example (type: (params: unknown) => unknown) if you delete the unknown. But I assume that's just for the sake of demonstration.

If you define the function foo as unknown the typescript compiler does not know which type foo has.

The good thing about this is that whoever uses this variable has to do a typecheck before the typescript compiler lets him pass.

const foo: unknown = (params: unknown): unknown => {
    return null;
}
// not working -> the typescript compiler says:
// This expression is not callable.
// Type '{}' has no call signatures.
foo('werwq');

So the one using that variable has to do a typecheck first and the typescript compiler let's the user of the variable use it as function.

if (typeof foo === "function") {
    foo('werwq');
}
// or even
switch (typeof foo) {
    case "function":
        foo("a");
}

Make sense this unknown type apply to the variables so it requires a type check — ANIK ISLAM SHOJIB, Nov 27 '20 at 16:44

variables type unknown not accessible typescript

1 Answers1