CODEWITHSUNDEEP
typescripttypescript2.0
Neo
Neo

Reputation: 4760

Why typescript is not throwing error even if an unwanted property is present in the interface?

Consider the following code block

interface Store {
    loading: boolean;
}

interface StoreMethod {
    (s: Store): Store;
}

export const createStore: StoreMethod = (store) => {
    return { ...store, working: false };
}


export const loadStore: StoreMethod = (store) => {
    return { loading: true, working: false };
}

I am expecting typescript to throw an error because in the return object is supposed to be Store which does not have property working. Am I missing something?

You can check the code in playground here.

Upvotes: 1

Views: 1043

Answers (1)

Trash Can
Trash Can

Reputation: 6814

You need to understand that typescript transpiles your code to plain javascript which has no concept of interfaces.

In typescript type A is compatible with type B if A has all of the properties that B does but A does not have to have the exact number of properties that B has, A has to at least have those properties that B does, aka, A could contain other properties which are not present in B. So this function

export const loadStore: StoreMethod = (store) => {
    return { loading: true, working: false };
}

The returned object clearly has a propert called loading which satifies the contract your Store imposes, so the returned object is considered compatible to Store, if you remove the loading property from the returned object, typescript will complain.

Interface in typescript is a design artifact, it is not present at runtime.

Upvotes: 4

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