no-unnecessary-type-assertion.md

description
Disallow type assertions that do not change the type of an expression.

🛑 This file is source code, not the primary documentation location! 🛑

See https://typescript-eslint.io/rules/no-unnecessary-type-assertion for documentation.

TypeScript can be told an expression is a different type than expected using as type assertions. Leaving as assertions in the codebase increases visual clutter and harms code readability, so it's generally best practice to remove them if they don't change the type of an expression. This rule reports when a type assertion does not change the type of an expression.

Examples

❌ Incorrect

const foo = 3;
const bar = foo!;

const foo = <3>3;

type Foo = 3;
const foo = <Foo>3;

type Foo = 3;
const foo = 3 as Foo;

const foo = 'foo' as const;

function foo(x: number): number {
  return x!; // unnecessary non-null
}

✅ Correct

const foo = <number>3;

const foo = 3 as number;

let foo = 'foo' as const;

function foo(x: number | undefined): number {
  return x!;
}

Options

typesToIgnore

With @typescript-eslint/no-unnecessary-type-assertion: ["error", { typesToIgnore: ['Foo'] }], the following is correct code:

type Foo = 3;
const foo: Foo = 3;

When Not To Use It

If you don't care about having no-op type assertions in your code, then you can turn off this rule.