Check if an object implements an interface at runtime with TypeScript

Question

I load a JSON configuration file at runtime, and use an interface to define its expected structure:

interface EngineConfig {     pathplanner?: PathPlannerConfig;     debug?: DebugConfig;     ... }  interface PathPlannerConfig {     nbMaxIter?: number;     nbIterPerChunk?: number;     heuristic?: string; }  interface DebugConfig {     logLevel?: number; }  ... 

This makes it convenient to access the various properties since I can use autocompletions etc.

Question: is there a way to use this declaration to check the correctness of the file I load? ie that I do not have unexpected properties?

Answer 1

There "is" a way, but you have to implement it yourself. It's called a "User Defined Type Guard" and it looks like this:

interface Test {     prop: number; }  function isTest(arg: any): arg is Test {     return arg && arg.prop && typeof(arg.prop) == 'number'; } 

Of course, the actual implementation of the isTest function is totally up to you, but the good part is that it's an actual function, which means it's testable.

Now at runtime you would use isTest() to validate if an object respects an interface. At compile time typescript picks up on the guard and treats subsequent usage as expected, i.e.:

let a:any = { prop: 5 };  a.x; //ok because here a is of type any  if (isTest(a)) {     a.x; //error because here a is of type Test } 

More in-depth explanations here: https://basarat.gitbook.io/typescript/type-system/typeguard