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Disallow calling a function with a value with type any.


该规则需要 类型信息 才能运行。

TypeScript 中的 any 类型是类型系统中危险的 "应急方案"。 使用 any 会禁用许多类型检查规则,通常最好仅作为最后手段或在构建代码原型时使用。

英: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.

尽管你的意图是好的,但 any 类型有时可能会泄漏到你的代码库中。 使用 any 类型参数调用函数会产生潜在的安全漏洞和错误来源。

英: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.

此规则不允许调用参数中包含 any 的函数。 这包括使用 any 类型元素作为函数参数来扩展数组或元组。

英:This rule disallows calling a function with any in its arguments. That includes spreading arrays or tuples with any typed elements as function arguments.

此规则还比较泛型类型参数类型,以确保你不会将泛型位置中的不安全 any 传递给需要特定类型的接收者。 例如,如果将 Set<any> 作为参数传递给声明为 Set<string> 的参数,则会出错。

英: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"
在线运行试试这个规则 ↗


declare function foo(arg1: string, arg2: number, arg3: string): void;

const anyTyped = 1 as any;

foo(anyTyped, 1, 'a');

const anyArray: any[] = [];

const tuple1 = ['a', anyTyped, 'b'] as const;

const tuple2 = [1] as const;
foo('a', ...tuple, anyTyped);

declare function bar(arg1: string, arg2: number, 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>());
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在某些情况下,规则允许将 any 参数传递到 unknown

英:There are cases where the rule allows passing an argument of any to unknown.

允许的 anyunknown 分配示例:

英: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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如果你的代码库有许多现有的 any 或不安全代码区域,则可能很难启用此规则。 在项目的不安全区域中,在增加类型安全性之前,跳过 no-unsafe-* 规则可能会更容易。 你可以考虑在这些特定情况下使用 ESLint 禁用注释,而不是完全禁用此规则。

英:If your codebase has many existing anys or areas of unsafe code, it may be difficult to enable this rule. It may be easier to skip the no-unsafe-* rules pending increasing type safety in unsafe areas of your project. You might consider using ESLint disable comments for those specific situations instead of completely disabling this rule.