Generating automatic generic types in Typescript without needing to explicitly declare the type

In the scenario where I have an interface containing two functions - one that returns a value, and another that uses the type of that value in the same interface - generics were initially used. However, every time a new object was created, the type had to be explicitly declared. But there is a way to achieve this without using generics. Consider the following example:

interface MyObject<T> {
    create(): T
    update(value: T): void // use value type defined in create method
}

Below is how we can define a new object utilizing the above interface:

let test: MyObject<string> = {
    create() {
        return "Hello"
    },
    update(value) {
        // do something with value
    },
}

Is it possible to write MyObject without <string>, since the value type has already been specified in the create method?

Answer №1

If you want the compiler to automatically determine the type string for the generic type parameter T, you will need to modify your code. This behavior only occurs when you call a generic function, so you'll have to refactor from using a type annotation (

const test: MyObject<string> = ...
) to calling a helper function (const test = toMyObject(...)).

Update for TS4.7+:

In TypeScript 4.7+, it seems that microsoft/TypeScript#48358 will introduce the ability for the compiler to contextualize type method parameters based on inferred type parameters from earlier defined properties. With this enhancement, the following code snippet will work:

const toMyObject = <T,>(o: MyObject<T>) => o;

let test = toMyObject({
    create() {
        return "Hello"
    },
    update(value) {
        value.toUpperCase()
    },
})

Note that the order of members in the object literal is crucial for this inference to work correctly. Therefore, make sure that T is inferred from create() before using it in update(). An incorrect order like the one below will cause an error:

let test2 = toMyObject({
    update(value) {
        value.toUpperCase() // error
    },
    create() {
        return "Hello"
    },
});

Playground link for TS4.7+


Previous answer for TS4.6-:

Additionally, if you need the compiler to infer the type of the callback parameter value in the update method, you should be aware of contextual typing. Unfortunately, in TypeScript, contextual type inference of callback parameters and generic type parameter inference might conflict when both rely on the same object. Refer to microsoft/TypeScript#38872 for more details.

To achieve both types of inference, split the input object into two parts within your helper function: one handling create for inferring the generic type parameter T, and the other dealing with update to allow inference of the callback parameter value's type.

This results in the following code:

const toMyObject = <T,>(
    create: () => T,
    update: (value: T) => void
): MyObject<T> => ({ create, update });

Let's try it out:

let test = toMyObject(
    () => { return "Hello" },
    (value) => { console.log(value.toUpperCase()); }
);
// let test: MyObject<string>

It appears to be working fine. The compiler determines that test is of type

MyObject<string></code as expected, and the callback parameter <code>value
is also inferred as string (as shown by the fact that value.toUpperCase() can be called without any errors in --strict mode).

Playground link to code for TS4.6-

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