Type definition in object literal in TypeScript

Update 2019-05-15 (Improved Code Pattern as Alternative)

After many years of using const and benefiting from more functional code, I would recommend against using the below in most cases. (When building objects, forcing the type system into a specific type instead of letting it infer types is often an indication that something is wrong).

Instead I would recommend using const variables as much as possible and then compose the object as the final step: 

const id = GetId();
const hasStarted = true;
...
const hasFinished = false;
...
return {hasStarted, hasFinished, id};
  • This will properly type everything without any need for explicit typing.
  • There is no need to retype the field names.
  • This leads to the cleanest code from my experience.
  • This allows the compiler to provide more state verification (for example, if you return in multiple locations, the compiler will ensure the same type of object is always returned - which encourages you to declare the whole return value at each position - giving a perfectly clear intention of that value).

Addition 2020-02-26

If you do actually need a type that you can be lazily initialized: Mark it is a nullable union type (null or Type). The type system will prevent you from using it without first ensuring it has a value.

In tsconfig.json, make sure you enable strict null checks:

"strictNullChecks": true

Then use this pattern and allow the type system to protect you from accidental null/undefined access:



const state = {
    instance: null as null | ApiService,
    // OR
    // instance: undefined as undefined | ApiService,

};

const useApi = () => {
    // If I try to use it here, the type system requires a safe way to access it

    // Simple lazy-initialization 
    const api = state?.instance ?? (state.instance = new ApiService());
    api.fun();

    // Also here are some ways to only access it if it has value:

    // The 'right' way: Typescript 3.7 required
    state.instance?.fun();

    // Or the old way: If you are stuck before Typescript 3.7
    state.instance && state.instance.fun();

    // Or the long winded way because the above just feels weird
    if (state.instance) { state.instance.fun(); }

    // Or the I came from C and can't check for nulls like they are booleans way
    if (state.instance != null) { state.instance.fun(); }

    // Or the I came from C and can't check for nulls like they are booleans 
    // AND I was told to always use triple === in javascript even with null checks way
    if (state.instance !== null && state.instance !== undefined) { state.instance.fun(); }
};

class ApiService {
    fun() {
        // Do something useful here
    }
}

Do not do the below in 99% of cases:

Update 2016-02-10 - To Handle TSX (Thanks @Josh)

Use the as operator for TSX.

var obj = {
    property: null as string
};

A longer example:

var call = {
    hasStarted: null as boolean,
    hasFinished: null as boolean,
    id: null as number,
};

Original Answer

Use the cast operator to make this succinct (by casting null to the desired type).

var obj = {
    property: <string> null
};

A longer example:

var call = {
    hasStarted: <boolean> null,
    hasFinished: <boolean> null,
    id: <number> null,
};

This is much better than having two parts (one to declare types, the second to declare defaults):

var callVerbose: {
    hasStarted: boolean;
    hasFinished: boolean;
    id: number;
} = {
    hasStarted: null,
    hasFinished: null,
    id: null,
};

You're pretty close, you just need to replace the = with a :. You can use an object type literal (see spec section 3.5.3) or an interface. Using an object type literal is close to what you have:

var obj: { property: string; } = { property: "foo" };

But you can also use an interface

interface MyObjLayout {
    property: string;
}

var obj: MyObjLayout = { property: "foo" };

Tags:

Typescript

Recent Posts