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no-unsafe-argument

Disallow calling a function with a value with type any.

💭

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>.

.eslintrc.cjs
module.exports = {
"rules": {
"@typescript-eslint/no-unsafe-argument": "error"
}
};
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Examples

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 Playground

There 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[]);
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Options

This rule is not configurable.

Resources