typescript - TypeScript 如何理解这种情况？

Question

当将一个unknown类型的变量赋值给另一个变量时，我们应该对unknown变量进行类型检查，但是TypeScript如何理解这个特定的条件呢？看起来它返回一个 boolean 但我认为它更复杂:

let first: unknown = 5;
let second: number;

const result = typeof first === 'number';

if(result){
    second = first; // error : Type 'unknown' is not assignable to type 'number'
}

if(typeof first === 'number'){
    second = first; // no errors
}

Answer 1

TypeScript added support for indirect expression type narrowing in TypeScript 4.4 after PR #44730 was added 2021 年 6 月。在介绍此功能的注释中，Anders Hejsberg 提到了此功能的局限性(强调并格式化了我的):

Narrowing through indirect-references occurs only when:

• the conditional-expression or discriminant-property access is declared in a const variable declaration with no type annotation, and...
• the reference being narrowed is one-of:
  • a const variable or
  • a readonly property or
  • a parameter for which there are no assignments in the function body.

在您的情况下，您的 let first:unknown = 5; 变量的赋值不能与此功能一起使用，因为它是 let 局部变量而不是 const 本地。

因此，如果您将代码更改为使用 constfirst 而不是 letfirst，那么它就可以工作:

const first: unknown = 5;
let second: number;

const result = ( typeof first === 'number' );

if( result ) {
    second = first; // OK!
}

if( typeof first === 'number' ) {
    second = first; // OK!
}

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