Here is my case:
/**
* Add types to the `add` function, so it:
* - accepts `a` parameter of type string or number
* - accepts `b` parameter of type string or number
* - returns type number when both `a` and `b` are of type number
* - returns type string when at least one param is of type string
*
* ⚠️ There is no need to modyfy the body of the function.
*/
function add(a,b) {
if (typeof a === 'number' && typeof b === 'number') {
return a b;
}
return String(a) String(b);
}
////////////////////////////////////////////////////////////////////////////////
//////////////////// TEST CASES - look but don't modify ////////////////////
////////////////////////////////////////////////////////////////////////////////
const sumA = add(1, 2);
const sumB = add(1, '2');
const sumC = add('1', 2);
const sumD = add('1', '2');
type cases = [
Expect<Equal<typeof sumA, number>>,
Expect<Equal<typeof sumB, string>>,
Expect<Equal<typeof sumC, string>>,
Expect<Equal<typeof sumD, string>>,
]
type Expect<T extends true> = T
type Equal<X, Y> =
(<T>() => T extends X ? 1 : 2) extends
(<T>() => T extends Y ? 1 : 2) ? true : false
No matter what I do (conditional return, type aliases) it does not allow me to pass through Expect cases.
Does anyone have an idea how to solve it?
CodePudding user response:
Overloads will help here:
function add(a: number, b: number): number
function add(a: string | number, b: string | number): string
function add(a: string | number, b: string | number): string | number {
if (typeof a === 'number' && typeof b === 'number') {
return a b;
}
return String(a) String(b);
}
CodePudding user response:
You can make use of Overloading
feature of JavaScript here.
Add overloaded definitions of the given function as follows
function add(a: number, b: number) : number
function add(a: string | number, b: string | number) : string
function add(a: string | number,b: string |number) : string | number {
if (typeof a === 'number' && typeof b === 'number') {
return a b;
}
return String(a) String(b);
}