Suggestion: Automatic type guard generation #15265

Imagine we want to deserialize some JSON object and then extract some information from it, this is a very common task. We expect the JSON data to be of a certain type, so wouldnt it be good if we could declare some type for our JSON:

interface Example {
    name: string;
    age: number;
    contactDetails: {
        address: string;
        phone: string
    }
}

and have an automatically generated type guard isExample(o : any): o is Example
like this:

function isExample(o) {
    return 'name' in o && typeof o.name === 'string' &&
               'age' in o && typeof o.age === 'number' &&
               'contactDetails' in o && typeof o.contactDetails === 'object' &&
                   'address' in o.contactDetails && typeof o.contactDetails.address === 'string' &&
                   'address' in o.phone && typeof o.contactDetails.phone === 'string';
}

So now I can do

let json = getJson()
if(isExample(json)){
   // use it as an Example, it definitely implements Example
} else {
  // throw some error
}

• caveat: I realize you cannot validate a function type at runtime like (s: string) => number, so just ignore those cases, and throw a compile error if someone tries to generate an automatic guard for a type that has a function type nested in it somewhere.
• caveat: You should only generate the type guard if it used, so there shouldn't be type guards generated for any type I define.

The syntax I would like for this is is<T>(o: any): o is T so i can use it like is<Example>(json)
what's good about it is that is is already a keyword.
And this would generate a type guard in javascript like is_Example(o){...} or something.