Disallow variable declarations from shadowing variables declared in the outer scope (no-shadow)

Rule Details

This rule extends the base eslint/no-shadow rule. It adds support for TypeScript's this parameters and global augmentation, and adds options for TypeScript features.

How to Use

{
  // note you must disable the base rule as it can report incorrect errors
  "no-shadow": "off",
  "@typescript-eslint/no-shadow": ["error"]
}

Options

See eslint/no-shadow options. This rule adds the following options:

interface Options extends BaseNoShadowOptions {
  ignoreTypeValueShadow?: boolean;
  ignoreFunctionTypeParameterNameValueShadow?: boolean;
}

const defaultOptions: Options = {
  ...baseNoShadowDefaultOptions,
  ignoreTypeValueShadow: true,
  ignoreFunctionTypeParameterNameValueShadow: true,
};

ignoreTypeValueShadow

When set to true, the rule will ignore the case when you name a type the same as a variable.

TypeScript allows types and variables to shadow one-another. This is generally safe because you cannot use variables in type locations without a typeof operator, so there's little risk of confusion.

Examples of correct code with { ignoreTypeValueShadow: true }:

type Foo = number;
const Foo = 1;

interface Bar {
  prop: number;
}
const Bar = 'test';

ignoreFunctionTypeParameterNameValueShadow

When set to true, the rule will ignore the case when you name a function type argument the same as a variable.

Each of a function type's arguments creates a value variable within the scope of the function type. This is done so that you can reference the type later using the typeof operator:

type Func = (test: string) => typeof test;

declare const fn: Func;
const result = fn('str'); // typeof result === string

This means that function type arguments shadow value variable names in parent scopes:

let test = 1;
type TestType = typeof test; // === number
type Func = (test: string) => typeof test; // this "test" references the argument, not the variable

declare const fn: Func;
const result = fn('str'); // typeof result === string

If you do not use the typeof operator in a function type return type position, you can safely turn this option on.

Examples of correct code with { ignoreFunctionTypeParameterNameValueShadow: true }:

const test = 1;
type Func = (test: string) => typeof test;

Taken with ❤️ from ESLint core

Attributes

• [ ] ✅ Recommended
• [ ] 🔧 Fixable
• [ ] 💭 Requires type information