Question

我在打字稿中缺少有关泛型的知识。假设以下示例：

type Caster<T> = (original: unknown) => T

const castA = <T>(original) => original as T
//                 ^^^ Parameter 'original' implicitly has an 'any' type.
const castB: Caster = <T>(original) => original as T
//            ^^^ Generic type 'Caster' requires 1 type argument(s).            
const castC: Caster<T> = (original) => original as T
//                 ^^^ Cannot find name 'T'.      ^^^

castA，castB和castC做同样的事情：它们强制转换为T类型的值（很丑，但符合示例）。

类型Caster<T>描述了应用程序unknown -> T。我假设我可以创建此类函数的通用实现，但是castA，castB和castC都会产生编译错误（请参见注释）。

相反，以下方法效果很好：

// OK - but it's not generic
const castString: Caster<string> = (original: unknown) => original as string
// OK - but no reference to generic "Caster" type
const castGeneric = <T>(original: unknown) => original as T

在我看来有点晦涩。我如何获得这种行为？它将非常有用（考虑工厂方法）

谢谢

Answer 1

const castC: Caster<T> = (original) => original as T

期望类型为string

const castC: Caster<string> = (original) => original as string

当您调用函数时，实际上是在使用泛型。因此，type Caster<T> = (original: unknown) => T使用泛型声明类型，而const castC: Caster<string> = (original) => original as string使用它

const castGeneric: Caster<T> = <T>(original: unknown) => original as T

Caster<T>是castGeneric的类型声明，因此它需要实际的类型

Answer 2

const castC:Caster<T> = ...

您需要给T指定某种类型然后castC将转到(original: unknown) => T

例如const castD:Caster<string>与constD:(original: unknown)=>string

打字稿通用功能

2 个答案: