Advanced TypeScript Exercises - Question 3

#typescript #javascript

Hello again. Today's question is about typing function with two arguments being union type. The goal is to block possibility to pass mixed types into arguments, so if the first argument is a number then second also needs to be number, in other words there is dependency between arguments which we need to write.

function f(a: string | number, b: string | number) {
    if (typeof a === 'string') {
        return a + ':' + b; // no error but b can be number!
    } else {
        return a + b; // error as b can be number | string
    }
}
f(2, 3); // correct usage
f(1, 'a'); // should be error
f('a', 2); // should be error
f('a', 'b') // correct usage

The whole code can be found in the playground

There is not one possibility of correct typing, can you solve this puzzle in many ways? Is it possible to type it without using type assertions? Post your answers in comments. Yes you can change the implementation also, the key is to have the same behavior + type safety. Have fun! Answer will be published soon!

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Top comments (7)

Rahul Kashyap • Feb 17 '20 • Edited

A quick idea -

function f<T extends string | number>(a: T, b: T) {
    if ( typeof a === "string" ) {
        return a + ':' + b; // no error but b can be number!
    } else {
        return a as number + (b as number); // error as b can be number | string
    }
}

f(2, 3); // correct usage
f(1, 'a'); // should be error
f('a', 2); // should be error
f('a', 'b') // correct usage

Playground link

Pragmatic Maciej • Feb 17 '20

Nice, thanks but your code still doesn't compile, can you make a version which compiles? - Playground

Rahul Kashyap • Feb 17 '20

Updated the answer :)

Pragmatic Maciej • Feb 17 '20 • Edited

Now the interface of the function works in reverse way, what is correct is incorrect 😉

Rahul Kashyap • Feb 17 '20 • Edited

If i understood it correct, we are not allowed to change implementation of the method, just introduce types to make it work, right ?

Pragmatic Maciej • Feb 17 '20

Yes, implementation of the value level can stay, but assertion like as X is a type level. Currently implementation cannot compile because of lack of proper typing, there is a way to solve it without any assertion that is why I left it as it is, so with compile error.

You can modify the implementation, the key is to have it working as original + typed safe.

Timon van Spronsen • Feb 18 '20 • Edited

Another great challenge! I thought it'd be rather straight forward, but I've been sitting on this for at least 30 minutes and I don't have a solution yet. I'll give it another go tomorrow :)

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Advanced TypeScript Exercises - Question 3

Top comments (7)

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