no-unsafe-argument
Disallow calling a function with a value with type
any
.
Extending "plugin:@typescript-eslint/recommended-type-checked"
in an ESLint configuration enables this rule.
This rule requires type information to run.
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.
Despite your best intentions, the any
type can sometimes leak into your codebase.
Calling a function with an any
typed argument creates a potential safety hole and source of bugs.
This rule disallows calling a function with any
in its arguments.
That includes spreading arrays or tuples with any
typed elements as function arguments.
This rule also compares generic type argument types to ensure you don't pass an unsafe any
in a generic position to a receiver that's expecting a specific type.
For example, it will error if you pass Set<any>
as an argument to a parameter declared as Set<string>
.
module.exports = {
"rules": {
"@typescript-eslint/no-unsafe-argument": "error"
}
};
Examples
- ❌ Incorrect
- ✅ Correct
declare function foo(arg1: string, arg2: number, arg3: string): void;
const anyTyped = 1 as any;
foo(...anyTyped);
foo(anyTyped, 1, 'a');
const anyArray: any[] = [];
foo(...anyArray);
const tuple1 = ['a', anyTyped, 'b'] as const;
foo(...tuple1);
const tuple2 = [1] as const;
foo('a', ...tuple, anyTyped);
declare function bar(arg1: string, arg2: number, ...rest: string[]): void;
const x = [1, 2] as [number, ...number[]];
foo('a', ...x, anyTyped);
declare function baz(arg1: Set<string>, arg2: Map<string, string>): void;
foo(new Set<any>(), new Map<any, string>());
Open in Playgrounddeclare function foo(arg1: string, arg2: number, arg3: string): void;
foo('a', 1, 'b');
const tuple1 = ['a', 1, 'b'] as const;
foo(...tuple1);
declare function bar(arg1: string, arg2: number, ...rest: string[]): void;
const array: string[] = ['a'];
bar('a', 1, ...array);
declare function baz(arg1: Set<string>, arg2: Map<string, string>): void;
foo(new Set<string>(), new Map<string, string>());
Open in PlaygroundThere are cases where the rule allows passing an argument of any
to unknown
.
Example of any
to unknown
assignment that are allowed:
declare function foo(arg1: unknown, arg2: Set<unknown>, arg3: unknown[]): void;
foo(1 as any, new Set<any>(), [] as any[]);
Open in PlaygroundOptions
This rule is not configurable.