Discover the inferred arguments of methods declared on an object in TypeScript

Question

Discover the inferred arguments of methods declared on an object in TypeScript

Assume I have methods that are defined in the following way:

const methods = {
    methodOne(a: string) { return a; },

    methodTwo(a: number) { return a; },

    methodThree() {}
} as const;

I am able to deduce the type of methods:

type MethodDefinitions = typeof methods;

Now, let's consider a scenario where I want to create a function that has the ability to execute any method from methods like this:

function doSomething<T extends keyof MethodDefinitions>(t: T, arg: Parameters<MethodDefinitions[T]>[0]) {
    const method = methods[t];
    method(arg);
}

In this case, I anticipate method to infer a specific type based on T and methods. However, TypeScript playground reveals that the type of method is

(a: never) => string | number | void

, and raises an error when attempting to call method with arg.

How can this issue be resolved without resorting to using any? One workaround is casting arg to never (method(arg as never)), but this solution may not be ideal.

Visit TS playground for further reference: link.

typescript generics typescript-generics type-inference

Answer 1

Answer №1

We are facing two main issues:

Firstly, the args variable is an array of arguments and cannot be passed directly when calling a function. There are a few alternative options available:

   method.apply(null, args);
   //OR
    method(...args);

Secondly, it should be noted that while TypeScript may not be flawless, your code appears to protect the input - although TypeScript itself may not recognize this. To address this, you can ensure enforcement by casting to any. The revised approach would look like this:

    (method as any).apply(null, args);
    (method as any)(...args);

Combining these changes, the entire function will now appear as follows:

function doStuff<T extends keyof MethodDefinitions>(t: T, ...args: Parameters<MethodDefinitions[T]>) {
    const method = methods[t];
    (method as any).apply(null, args);
    //(method as any)(...args); - alternatively, you could opt for this method if preferred
}

To see a working example in action, visit the following Playground link: Playground

Answer 2

We are facing two main issues:

Firstly, the args variable is an array of arguments and cannot be passed directly when calling a function. There are a few alternative options available:

   method.apply(null, args);
   //OR
    method(...args);

Secondly, it should be noted that while TypeScript may not be flawless, your code appears to protect the input - although TypeScript itself may not recognize this. To address this, you can ensure enforcement by casting to any. The revised approach would look like this:

    (method as any).apply(null, args);
    (method as any)(...args);

Combining these changes, the entire function will now appear as follows:

function doStuff<T extends keyof MethodDefinitions>(t: T, ...args: Parameters<MethodDefinitions[T]>) {
    const method = methods[t];
    (method as any).apply(null, args);
    //(method as any)(...args); - alternatively, you could opt for this method if preferred
}

To see a working example in action, visit the following Playground link: Playground

Discover the inferred arguments of methods declared on an object in TypeScript

Answer №1

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