Get inferred type for a generic parameter in TypeScript

config.toml

@@ -63,7 +63,7 @@ images = [""]
 #
 homeSubtitle = [
   "Hi! I'm a Software Engineer, an avid Linux user, an enthusiast of many programming languages, a <a href='/recipes'>home cook</a>, and an amateur gardener.",
-  "My interests include building web and mobile applications, both front and back end. Over the years I learned and used many programming languages and technologies, including JavaScript, TypeScript, React, React Native, Python, Java, C, C++, Clojure, Rust, and Haskell. Pretty much everthing I've worked on is open source and available on <a href='https://github.com/SeriousBug/'>my Github page</a>.",
+  "My interests include building web and mobile applications, both front and back end. Over the years I learned and used many programming languages and technologies, including JavaScript, TypeScript, React, React Native, Python, Java, C, C++, Clojure, Rust, and Haskell. Pretty much everything I've worked on is open source and available on <a href='https://github.com/SeriousBug/'>my Github page</a>.",
 ]
 
 # Set a background for the homepage

content/posts/2022.01.28.typescript-get-inferred-type.md

@@ -0,0 +1,79 @@
+---
+title: "Get inferred type for a generic parameter in TypeScript"
+date: 2023-01-28T14:50:54-05:00
+toc: false
+images:
+tags:
+  - dev
+  - typescript
+---
+
+Have you used [Zod](https://zod.dev/)? It's a very cool TypeScript library for
+schema validation. Compared to alternatives like Joi, one of the biggest
+strenghts of Zod is that it can do type inference. For example,
+
+```ts
+const PersonSchema = z.object({
+  name: z.string(),
+  age: z.number(),
+});
+
+type Person = z.infer<typeof PersonSchema>;
+// This is equivalent to { name: string; age: number; }
+```
+
+Now I was recently working on a database client, where I'm using a validator
+function to ensure the data on the database matches what the client expects. I
+then take advantage of TypeScript's type inference so the type of everything
+matches. It looks like this:
+
+```ts
+class Database<T> {
+  constructor(validator: (input: unknown) => T);
+
+  function get(key: string): T | undefined { /* ... */ }
+  function put(key: string, data: T) { /* ... */ }
+}
+
+// Note I didn't have to specify the type parameter,
+// TypeScript infers it from the validator argument
+const PersonDB = new Database(PersonSchema.parse);
+```
+
+At this point I started to wonder, could I do something similar to what Zod does
+and get the inferred type for the objects that are stored in the database? While
+this is not required in this example above since I could get the type from Zod,
+the validator function doesn't necessarily have to be implemented with Zod.
+
+After reading through Zod's codebase, I found the trick they use, and it's very
+simple. Let's see it:
+
+```ts
+class Database<T> {
+  readonly _output!: T;
+
+  constructor(validator: (input: unknown) => T);
+
+  function get(key: string): T | undefined { /* ... */ }
+  function put(key: string, data: T) { /* ... */ }
+}
+
+type EntityOf<D extends Database<any>> = D["_output"];
+
+const PersonDB = new Database(PersonSchema.parse);
+
+type Person = EntityOf<typeof PersonDB>;
+```
+
+This is surprisingly simple. We add a property `_output` to the class, which has
+the inferred type. We can then get the type through that property with
+`D["_output"]`. The `!` in the definition of the property is there because we
+never actually set any value for `_output`. TypeScript normally will detect and
+warn us that we did not set `_output`, but the exclamation point suppresses
+that.
+
+This is not without drawbacks of course, because the `_output` property will be
+visible in the instances of the class. We can't hide the property with `private`
+because TypeScript won't let us look it up in `EntityOf` if we do so. So the
+best we can do is document the fact that this should not be used, and throw in
+the prefix so it stands out from regular properties.