no-unsafe-return

Disallow returning a value with type any from a function.

✅

在 ESLint 配置中扩展"plugin:@typescript-eslint/recommended-type-checked" 可启用此规则。

💭

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。从函数返回 any 类型的值会在代码库中创建潜在的类型安全漏洞和错误来源。

英：Despite your best intentions, the any type can sometimes leak into your codebase. Returning an an any-typed value from a function creates a potential type safety hole and source of bugs in your codebase.

此规则不允许从函数返回 any 或 any[]。

英：This rule disallows returning any or any[] from a function.

此规则还比较泛型类型参数类型，以确保你不会在泛型位置向需要特定类型的函数返回不安全的 any。例如，如果从声明为返回 Set<string> 的函数返回 Set<any>，则会出错。

英：This rule also compares generic type argument types to ensure you don't return an unsafe any in a generic position to a function that's expecting a specific type. For example, it will error if you return Set<any> from a function declared as returning Set<string>.

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

在线运行试试这个规则 ↗

示例

❌ 不正确
✅ 正确

function foo1() {
  return 1 as any;
}
function foo2() {
  return Object.create(null);
}
const foo3 = () => {
  return 1 as any;
};
const foo4 = () => Object.create(null);

function foo5() {
  return [] as any[];
}
function foo6() {
  return [] as Array<any>;
}
function foo7() {
  return [] as readonly any[];
}
function foo8() {
  return [] as Readonly<any[]>;
}
const foo9 = () => {
  return [] as any[];
};
const foo10 = () => [] as any[];

const foo11 = (): string[] => [1, 2, 3] as any[];

// generic position examples
function assignability1(): Set<string> {
  return new Set<any>([1]);
}
type TAssign = () => Set<string>;
const assignability2: TAssign = () => new Set<any>([true]);
Open in Playground

function foo1() {
  return 1;
}
function foo2() {
  return Object.create(null) as Record<string, unknown>;
}

const foo3 = () => [];
const foo4 = () => ['a'];

function assignability1(): Set<string> {
  return new Set<string>(['foo']);
}
type TAssign = () => Set<string>;
const assignability2: TAssign = () => new Set(['foo']);
Open in Playground

在某些情况下，规则允许将 any 返回到 unknown。

英：There are cases where the rule allows to return any to unknown.

允许的 any 到 unknown 返回示例：

英：Examples of any to unknown return that are allowed:

function foo1(): unknown {
  return JSON.parse(singleObjString); // Return type for JSON.parse is any.
}

function foo2(): unknown[] {
  return [] as any[];
}
Open in Playground

何时不使用它

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.

选项

该规则不可配置。

示例​

何时不使用它​

相关​

选项​

资源​

示例

何时不使用它

相关

选项

资源