typescript/no-unnecessary-type-constraint Suspicious

What it does

Disallow unnecessary constraints on generic types.

Why is this bad?

Generic type parameters (<T>) in TypeScript may be "constrained" with an extends keyword. When no extends is provided, type parameters default a constraint to unknown. It is therefore redundant to extend from any or unknown.

Examples

Examples of incorrect code for this rule:

interface FooAny<T extends any> {}
interface FooUnknown<T extends unknown> {}

type BarAny<T extends any> = {};
type BarUnknown<T extends unknown> = {};

const QuuxAny = <T extends any>() => {};

function QuuzAny<T extends any>() {}

class BazAny<T extends any> {
  quxAny<U extends any>() {}
}

Examples of correct code for this rule:

interface Foo<T> {}

type Bar<T> = {};

const Quux = <T>() => {};

function Quuz<T>() {}

class Baz<T> {
  qux<U>() {}
}

How to use

To enable this rule in the CLI or using the config file, you can use:

CLI

oxlint --deny typescript/no-unnecessary-type-constraint

Config (.oxlintrc.json)

{
  "rules": {
    "typescript/no-unnecessary-type-constraint": "error"
  }
}

References

• Rule Source