error[no-slow-types]: super class expression was too complex

I am using Effect, and when I try to lint the codebase I get errors about super classes

x deno lint
error[no-slow-types]: super class expression was too complex
  |
7 |   extends Schema.TaggedError<FirebaseError>("FirebaseError")("FirebaseError", {
  |           ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
  |
8 |     error: Schema.Unknown,
  | ^^^^^^^^^^^^^^^^^^^^^^^^^^
  |
9 |   }) {}
  | ^^^^ this is the superclass expression
  |
  = hint: extract the superclass expression into a variable

  info: fast check was unable to infer the type of the superclass expression
  docs: https://jsr.io/go/slow-type-unsupported-super-class-expr

x deno lint
error[no-slow-types]: super class expression was too complex
  |
7 |   extends Schema.TaggedError<FirebaseError>("FirebaseError")("FirebaseError", {
  |           ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
  |
8 |     error: Schema.Unknown,
  | ^^^^^^^^^^^^^^^^^^^^^^^^^^
  |
9 |   }) {}
  | ^^^^ this is the superclass expression
  |
  = hint: extract the superclass expression into a variable

  info: fast check was unable to infer the type of the superclass expression
  docs: https://jsr.io/go/slow-type-unsupported-super-class-expr

The code in question is this:

export class FirebaseError
  extends Schema.TaggedError<FirebaseError>("FirebaseError")("FirebaseError", {
    error: Schema.Unknown,
  }) {}

export class FirebaseError
  extends Schema.TaggedError<FirebaseError>("FirebaseError")("FirebaseError", {
    error: Schema.Unknown,
  }) {}

After reading a bit about slow types I am guessing that the type export is too complex to be inferred, and could slow down type checking the code base. The solution is to type the exported things, but I don't know how. In the documentation it says that I should do this: For super class expressions, evaluate the expression and assign it to an intermediate type:

  interface ISuperClass {}

  function getSuperClass() {
    return class SuperClass implements ISuperClass {};
  }

- export class MyClass extends getSuperClass() {}
+ const SuperClass: ISuperClass = getSuperClass();
+ export class MyClass extends SuperClass {}

  interface ISuperClass {}

  function getSuperClass() {
    return class SuperClass implements ISuperClass {};
  }

- export class MyClass extends getSuperClass() {}
+ const SuperClass: ISuperClass = getSuperClass();
+ export class MyClass extends SuperClass {}

but I don't know how to apply that to Schema.TaggedError. Also, it complains about the same thing for Effect.service

export class Firebase extends Effect.Service<Firebase>()("Firebase", {
  effect: Effect.gen(function* () {}){}

export class Firebase extends Effect.Service<Firebase>()("Firebase", {
  effect: Effect.gen(function* () {}){}

I can ignore slow types in the config, but that means that by using Effect in packages I am creating slow types all over the place. Any ideas?

2 Replies

marvinh.•4h ago

Can you file an issue for that https://github.com/denoland/deno/issues ?

GitHub

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mabasicOP•2h ago

Done: https://github.com/denoland/deno/issues/31290

error[no-slow-types]: super class expression was too complex

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