Interface type check with Typescript

Question

This question is the direct analogon to Class type check with TypeScript

I need to find out at runtime if a variable of type any implements an interface. Here\'s my

Answer 1

Type guards in Typescript:

TS has type guards for this purpose. They define it in the following manner:

Some expression that performs a runtime check that guarantees the type in some scope.

This basically means that the TS compiler can narrow down the type to a more specific type when it has sufficient information. For example:

function foo (arg: number | string) {
    if (typeof arg === 'number') {
        // fine, type number has toFixed method
        arg.toFixed()
    } else {
        // Property 'toFixed' does not exist on type 'string'. Did you mean 'fixed'?
        arg.toFixed()
        // TSC can infer that the type is string because 
        // the possibility of type number is eliminated at the if statement
    }
}

To come back to your question, we can also apply this concept of type guards to objects in order to determine their type. To define a type guard for objects, we need to define a function whose return type is a type predicate. For example:

interface Dog {
    bark: () => void;
}

// The function isDog is a user defined type guard
// the return type: 'pet is Dog' is a type predicate, 
// it determines whether the object is a Dog
function isDog(pet: object): pet is Dog {
  return (pet as Dog).bark !== undefined;
}

const dog: any = {bark: () => {console.log('woof')}};

if (isDog(dog)) {
    // TS now knows that objects within this if statement are always type Dog
    // This is because the type guard isDog narrowed down the type to Dog
    dog.bark();
}