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no-unsafe-argument

Disallows calling a function with a value with type any.

Despite your best intentions, the any type can sometimes leak into your codebase. Call a function with any typed argument are not checked at all by TypeScript, so it creates a potential safety hole, and source of bugs in your codebase.

Attributes​

  • Included in configs
    • ✅ Recommended
    • 🔒 Strict
  • Fixable
    • 🔧 Automated Fixer
    • 🛠 Suggestion Fixer
  • 💭 Requires type information

Rule Details​

This rule disallows calling a function with any in its arguments, and it will disallow spreading any[]. This rule also disallows spreading a tuple type with one of its elements typed as any. This rule also compares the argument's type to the variable's type 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 assign Set<any> to an argument declared as Set<string>.

Examples of code for this rule:

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>());

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<unkown>, arg3: unknown[]): void;
foo(1 as any, new Set<any>(), [] as any[]);

Options​

// .eslintrc.json
{
  "rules": {
    "@typescript-eslint/no-unsafe-argument": "error"
  }
}

This rule is not configurable.