英文:
How can TypeScript understand this condition?
问题
当将类型为unknown的变量分配给另一个变量时，我们应该经过unknown变量的类型检查，但TypeScript如何理解这种特定情况呢？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; // 错误：类型'unknown'不能赋值给类型'number'
}
if(typeof first === 'number'){
second = first; // 没有错误
}
如果您有任何其他翻译需求，请告诉我。
英文:
when assigning a variable of type unknown to another variable, we should go through the type checking of the unknown variable but how can TypeScript understand this specific condition? looks like it's returning a boolean but I think it's more complicated :
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
}
答案1
得分: 5
TypeScript在2021年6月的PR＃44730之后，增加了对间接表达式类型缩小的支持。在介绍此功能的说明中，Anders Hejsberg提到了此功能的限制（重点和格式由我添加）：
> 只有在以下情况下，通过_间接引用_进行缩小才会发生：
> * _条件表达式_或_鉴别属性_访问是在没有类型注释的const变量声明中声明的，并且...
> * 被缩小的引用是以下之一：
> * const变量 或
> * readonly属性 或
> * 函数体中没有赋值的参数。
在您的案例中，您的let first: unknown = 5;变量的赋值不能使用这个功能，因为它是一个let局部变量，而不是一个const局部变量。
因此，如果您将代码更改为使用const first而不是let first，那么它就能正常工作：
const first: unknown = 5;
let second: number;
const result = ( typeof first === 'number' );
if( result ) {
second = first; // OK!
}
if( typeof first === 'number' ) {
second = first; // OK!
}
英文:
TypeScript added support for indirect expression type narrowing in TypeScript 4.4 after PR #44730 was added in June 2021. In the notes introducing this feature, Anders Hejsberg mentions the limitations of this feature (emphasis and formatting mine):
> 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.
In your case your let first: unknown = 5; variable's assign cannot be used with this feature because it's a let local variable instead of a const local.
So if you change the code to use const first instead of let first, then it works:
const first: unknown = 5;
let second: number;
const result = ( typeof first === 'number' );
if( result ) {
second = first; // OK!
}
if( typeof first === 'number' ) {
second = first; // OK!
}
通过集体智慧和协作来改善编程学习和解决问题的方式。致力于成为全球开发者共同参与的知识库，让每个人都能够通过互相帮助和分享经验来进步。
评论