Union types can be used to constrain generic type parameters in Typescript

Question

Union types can be used to constrain generic type parameters in Typescript

I'm working on a function

 uniqueIds(first: any[], second: any[]): number[] {
        let prop = first[0] instanceof Owner ? "OwnerId"
            : "BankId";
        return _.unique(
            _.map(
                first,
                o => o[prop]
            )
            .concat(
                _.map(
                    second,
                    o => o[prop]
                )
            )
        ).filter(o => o);
    }

This function is specifically designed to work with either Bank or Owner types, and nothing else. These two types do not share any interface or inheritance chain.
My current implementation might not be the most elegant solution for this scenario. Is there a better way to achieve something like

uniqueIds<T> where T Bank | Owner

without having to stick with my current approach?

typescript typescript-generics

Answer 1

Answer №1

If you want to declare a function that accepts either Owner[] or Bank[], you can define it as follows:

function findUniqueIds<T extends Owner[] | Bank[]>(first: T, second: T) { ... }

However, TypeScript may raise a noImplicitAny error because it cannot verify that o[prop] is always a number. To address this, consider passing the property name along with the arrays:

function findUniqueIds<K extends string, T extends {[P in K]: number}>(
    prop: K, first: T[], second: T[]): number[] {
    return _.unique(
        _.map(
            first,
            o => o[prop]
        )
        .concat(
            _.map(
                second,
                o => o[prop]
            )
        )
    ).filter(o => o);
}

Answer 2

If you want to declare a function that accepts either Owner[] or Bank[], you can define it as follows:

function findUniqueIds<T extends Owner[] | Bank[]>(first: T, second: T) { ... }

However, TypeScript may raise a noImplicitAny error because it cannot verify that o[prop] is always a number. To address this, consider passing the property name along with the arrays:

function findUniqueIds<K extends string, T extends {[P in K]: number}>(
    prop: K, first: T[], second: T[]): number[] {
    return _.unique(
        _.map(
            first,
            o => o[prop]
        )
        .concat(
            _.map(
                second,
                o => o[prop]
            )
        )
    ).filter(o => o);
}

Union types can be used to constrain generic type parameters in Typescript

Answer №1

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