Declaring type of an object inside a function call?

Question

In TypeScript, I can do this:

function someFunction(options:any) {
    // Do something
}

interface MyOptions {
    userId: number;
    verbose: boolean;
}

const options: MyOptions = {
    userId: 12345,
    doesntExist: false, // Error
}

someFunction(options);

And that will give me a compilation error on the "doesntExist" property.

The problem is when I do the same thing without creating an intermediate variable:

someFunction({
    userId: 12345,
    doesntExist: false, // no error
});

In that case, since someFunction accepts any as a parameter, type checking is not enforced. Even if I do this, the compiler doesn't complain:

someFunction({
    userId: 12345,
    doesntExist: false, // no error
} as MyOptions);

So I'm wondering is there any way to enforce type checking without creating the intermediate variable?

Answer 1

What I was wondering is if there's a syntax that allows me to say "while the function takes 'any' as parameter, I declare that the object I'm passing is of type T"

What you are describing is generics.

function someFunction<T>(options: T) {
  // Do something
}

If you just call the function without specifying the generic T then you won't get an error because typescript will assume that T is the type for whatever argument you provided.

But you can manually enter the generic as MyOptions and you will get an error when your argument isn't right.

someFunction<MyOptions>({
  userId: 12345,
  doesntExist: false // error
})

Here we get the error Argument of type '{ userId: number; doesntExist: boolean; }' is not assignable to parameter of type 'MyOptions'. Object literal may only specify known properties, and 'doesntExist' does not exist in type 'MyOptions'.

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