The function type is not as strict as the return type in terms of enforcement

I am working on defining a function type that enforces the return type (Object) to have the exact object properties specified in the return type definition.

Despite my efforts, the compiler is not enforcing strict adherence to the returned object properties and allows additional properties that are not defined in the return type.

interface Obj {
  foo: string;
}

type Func = () => Obj;

const fn: Func = () => {
  return {
    foo: 'bar',
    blah: '', // the compiler does not raise an error
  };
};

Example in TS playground

However, if I explicitly specify the return type as Obj in the fn function, the compiler correctly flags any extra properties.

interface Obj {
  foo: string;
}

const fn = (): Obj => {
  return {
    foo: 'bar',
    blah: '', // the compiler raises an error
  };
};

Example in TS playground

Could someone provide insight into why TypeScript handles these scenarios differently? Is there a way to maintain return type strictness when using a function type?

Answer №1

When looking at the first example, the return type of the anonymous function is dynamically defined as { foo: string; blah: string; }. This particular type is considered to be compatible with the type Obj, since it includes all the necessary properties of Obj (and more, although meeting the basic requirements is sufficient for compatibility, as "structural typing is a way of relating types based solely on their members", according to https://www.typescriptlang.org/docs/handbook/type-compatibility.html).

In contrast, in the second example, the signature specifies the return type of the anonymous function itself. Therefore, the returned object must be an exact match for type Obj. Just like you cannot write

let obj: Obj = { foo:'bar', blah:'' }
, this code will not compile because there is no room for compatibility checking in this scenario where the type is explicitly stated.

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