This piece of code encapsulates the essence of what I'm trying to achieve more effectively than words:
function A(a: string): string;
function A(a: number): number;
function A(a: any) { return a; }
function B<T extends number | string>(arg: T): T {
return A(arg); // TS error: No overload matches this call.
}
Is it possible to use A(num), A(str), but not A(num | string) in this scenario?
Do you have any suggestions on how to resolve this issue?
Have you considered implementing a generic method for A?
Perhaps something along these lines:
function A<T>(a: T): T { return a; }
function B<T extends number | string>(arg: T): T {
return A(arg); // TypeScript error: No overload matches this call.
}
One common misconception is that when you declare T extends number | string, it implies that T can only be a number or a string. However, there are actually subtypes of number | string that do not fall under the categories of number or string.
type U = 1 | 'one' | 2 | 'two';
declare const arg: U;
B(arg); // This is valid because `U` is within the limits of `number | string`
A(arg); // ERROR: type `U`
// does not fit into the category of `number` or `string`
To successfully incorporate the signature of A without any overloads, you must follow this implementation:
function A<T extends number | string>(a: T): T { return a; }
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