---
description: 'Disallow variable declarations from shadowing variables declared in the outer scope.'
---

> 🛑 This file is source code, not the primary documentation location! 🛑
>
> See **https://typescript-eslint.io/rules/no-shadow** for documentation.

## Examples

This rule extends the base [`eslint/no-shadow`](https://eslint.org/docs/rules/no-shadow) rule.
It adds support for TypeScript's `this` parameters and global augmentation, and adds options for TypeScript features.

## Options

This rule adds the following options:

```ts
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 }`:

```ts
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:

```ts
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:

```ts
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 }`:

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

## FAQ

### Why does the rule report on enum members that share the same name as a variable in a parent scope?

Reporting on this case isn't a bug - it is completely intentional and correct reporting! The rule reports due to a relatively unknown feature of enums - enum members create a variable within the enum scope so that they can be referenced within the enum without a qualifier.

To illustrate this with an example:

```ts
const A = 2;
enum Test {
  A = 1,
  B = A,
}

console.log(Test.B);
// what should be logged?
```

Naively looking at the above code, it might look like the log should output `2`, because the outer variable `A`'s value is `2` - however, the code instead outputs `1`, which is the value of `Test.A`. This is because the unqualified code `B = A` is equivalent to the fully-qualified code `B = Test.A`. Due to this behavior, the enum member has **shadowed** the outer variable declaration.