断言函数用于函数表达式时，ts 报错，

[liuben-team]

文档地址

// 写法一
const assertIsNumber = (value: unknown): asserts value is number => {
  if (typeof value !== 'number')
    throw Error('Not a number');
};
assertIsNumber(1)

tsconfig.json ("typescript": "^5.6.2")

{
  "compilerOptions": {
    "allowImportingTsExtensions": true,
    "allowUnreachableCode": false,
    "allowUnusedLabels": false,
    "alwaysStrict": true,
    "exactOptionalPropertyTypes": true,
    "module": "NodeNext",
    "moduleResolution": "nodenext",
    "noEmit": true,
    "noFallthroughCasesInSwitch": true,
    "noImplicitAny": true,
    "noImplicitOverride": true,
    "noImplicitReturns": true,
    "noImplicitThis": true,
    "noPropertyAccessFromIndexSignature": false,
    "noUncheckedIndexedAccess": false,
    "noUnusedLocals": true,
    "noUnusedParameters": true,
    "strict": true,
    "strictBindCallApply": true,
    "strictFunctionTypes": true,
    "strictNullChecks": true,
    "strictPropertyInitialization": true,
    "target": "ESNext",
    "useUnknownInCatchVariables": false
  }
}

[liuben-team]

网上有一些说法是：为了确保类型断言函数在使用箭头函数定义时能够正确工作，我们需要确保类型断言函数的定义包含显式的类型注解。并给出下面的解决方案，报错依然存在

const n: unknown = 1;
assertIsNumber(n)

[ruanyf]

https://www.typescriptlang.org/play/?#code/MYewdgzgLgBAhhCBTATlAkhAcgVwLYBGqMAvDABQBQMMAbnADY5IBcOYA1mCAO5iUBKFgmRoIdRsxgBLcWHxEUpAHwwA3tRkAzClACeAByQgd9JkhgBCEmQDk8wqlsDNNKAAsUvGAFEUXlHJbLBBYOBgHRWcAbkoAX2igA

[liuben-team]

https://www.typescriptlang.org/play/?#code/MYewdgzgLgBAhhCBTATlAkhAcgVwLYBGqMAvDABQBQMMAbnADY5IBcOYA1mCAO5iUBKFgmRoIdRsxgBLcWHxEUpAHwwA3tRkAzClACeAByQgd9JkhgBCEmQDk8wqlsDNNKAAsUvGAFEUXlHJbLBBYOBgHRWcAbkoAX2igA

[ruanyf]

参考 microsoft/TypeScript#33622 (comment) ，删除写法一。