no-shadow
Disallow variable declarations from shadowing variables declared in the outer scope.
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
module.exports = {
"rules": {
// Note: you must disable the base rule as it can report incorrect errors
"no-shadow": "off",
"@typescript-eslint/no-shadow": "error"
}
};
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';
Open in PlaygroundignoreFunctionTypeParameterNameValueShadow
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;
Open in PlaygroundFAQ
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:
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.
Resources
Taken with ❤️ from ESLint core