validation - Typescript: Declare function that takes parameter of exact interface type, and not one of its derived types -

interface {     a: number; }  interface ii extends {     b: number; }  function f(arg: i) : void {     // arg without trimming properties (logical error)     console.log(arg); }  const obj: ii = { a:4, b:3 }; f(obj); 

what want make function f accept objects of type i , not type ii or other derived interface

you cannot achieve in typescript, in general, in languages cannot make such constraint. 1 principle of object oriented programming can pass derived class base class expected. can perform runtime check , if find members don't expect, can throw error. compiler not achieve this.