Error 2322: Troubleshooting Typescript arrow functions overloads issues

Question

Error 2322: Troubleshooting Typescript arrow functions overloads issues

Everything seems to be working well with the code below, except for an error that occurs with the resolve constant.

const resolve: Resolve Type '(param: "case 1" | "case 2" | "case 3") => boolean | "string" | 1000' is not assignable to type 'Resolve'.(2322)

// Overloads
type Resolve = {
    (): false;
    (param: 'case 1'): string;
    (param: 'case 2'): number;
    (param: 'case 3'): true;
};

const resolve: Resolve = (param) => {
    switch (param) {
        case 'case 1':
            return 'string';
        case 'case 2':
            return 1000;
        case 'case 3':
            return true;
        default:
            return false;
    }
};
    
const result = {
    first: resolve('case 1'),
    second: resolve('case 2'),
    third: resolve('case 3'),
    none: resolve()
};

Do you have any suggestion on how to fix this issue?

typescript function arrow-functions constructor-overloading

Answer 1

Answer №1

It's a common scenario where the implementation signature of an overloaded function may not completely align with the overload signatures. TypeScript employs relaxed rules for compatibility when using overload syntax.

In your situation, since you are not utilizing overload syntax, a type assertion needs to be used (after ensuring that the function operates correctly for the overloads):

const resolve = ((param?: "case 1" | "case 2" | "case 3"): string | number | true | false => {
    switch (param) {
        case "case 1":
            return "string";
        case "case 2":
            return 1000;
        case "case 3":
            return true;
        default:
            return false;
    }
}) as Resolve;

(I've also included a type annotation for param and a return type annotation for precision, although being lenient in the implementation signature is common.)

Playground example

For what it's worth, consider using function overload syntax instead (playground link):

function resolve(param: "case 1"): string;
function resolve(param: "case 2"): number;
function resolve(param: "case 3"): true;
function resolve(): false;
function resolve(param?: "case 1" | "case 2" | "case 3"): string | number | true | false {
    switch (param) {
        case "case 1":
            return "string";
        case "case 2":
            return 1000;
        case "case 3":
            return true;
        default:
            return false;
    }
};

type Resolve = typeof resolve;

...however, there are scenarios where defining the type without actual implementation becomes necessary (such as in a callback type for an API).

Answer 2

It's a common scenario where the implementation signature of an overloaded function may not completely align with the overload signatures. TypeScript employs relaxed rules for compatibility when using overload syntax.

In your situation, since you are not utilizing overload syntax, a type assertion needs to be used (after ensuring that the function operates correctly for the overloads):

const resolve = ((param?: "case 1" | "case 2" | "case 3"): string | number | true | false => {
    switch (param) {
        case "case 1":
            return "string";
        case "case 2":
            return 1000;
        case "case 3":
            return true;
        default:
            return false;
    }
}) as Resolve;

(I've also included a type annotation for param and a return type annotation for precision, although being lenient in the implementation signature is common.)

Playground example

For what it's worth, consider using function overload syntax instead (playground link):

function resolve(param: "case 1"): string;
function resolve(param: "case 2"): number;
function resolve(param: "case 3"): true;
function resolve(): false;
function resolve(param?: "case 1" | "case 2" | "case 3"): string | number | true | false {
    switch (param) {
        case "case 1":
            return "string";
        case "case 2":
            return 1000;
        case "case 3":
            return true;
        default:
            return false;
    }
};

type Resolve = typeof resolve;

...however, there are scenarios where defining the type without actual implementation becomes necessary (such as in a callback type for an API).

Error 2322: Troubleshooting Typescript arrow functions overloads issues

Answer №1

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