typescript/no-explicit-any Restriction
What it does
Disallows explicit use of the any type.
Why is this bad?
The any type in TypeScript is a dangerous "escape hatch" from the type system. Using any disables many type checking rules and is generally best used only as a last resort or when prototyping code. This rule reports on explicit uses of the any keyword as a type annotation.
TypeScript's
--noImplicitAnycompiler option prevents an impliedany, but doesn't preventanyfrom being explicitly used the way this rule does.
Examples
Examples of incorrect code for this rule:
typescript
const age: any = "seventeen";
const ages: any[] = ["seventeen"];
const ages: Array<any> = ["seventeen"];
function greet(): any {}
function greet(): any[] {}
function greet(): Array<any> {}
function greet(): Array<Array<any>> {}
function greet(param: Array<any>): string {}
function greet(param: Array<any>): Array<any> {}Examples of correct code for this rule:
typescript
const age: number = 17;
const ages: number[] = [17];
const ages: Array<number> = [17];
function greet(): string {}
function greet(): string[] {}
function greet(): Array<string> {}
function greet(): Array<Array<string>> {}
function greet(param: Array<string>): string {}
function greet(param: Array<string>): Array<string> {}Options
ignoreRestArgs
A boolean to specify if arrays from the rest operator are considered ok. false by default.
fixToUnknown
Whether to enable auto-fixing in which the any type is converted to the unknown type. false by default.
How to use
To enable this rule in the CLI or using the config file, you can use:
bash
oxlint --deny typescript/no-explicit-anyjson
{
"rules": {
"typescript/no-explicit-any": "error"
}
}