Extending a string in TypeScript

Question

When using generics in TypeScript, you sometimes see a type Parameter such as:

T extends string

Isn’t this the same as using string directly? Can you subclass string? What would this be good for?

Answer 1

下面是一个简单的示例，说明如何扩展字符串以及对类型进行硬编码会破坏它。

type Foo = string & Partial<{ meta: string }>

const foo: Foo = 'foo';
foo.meta = 'this string is important';

function hardCoded(x: string): string {
  return x;
}

hardCoded(foo).meta;  // this is an error

function generic<X extends string>(x: X): X {
  return x;
}
generic(foo).meta;  // but this works

Answer 2

type narrowedString =  "foo" | "bar"

// type ExtendsString = true
type ExtendsString = "foo" extends string ? true : false 

"foo" and "bar" extend both the string type. For example that is useful, when you want to define enum like data structures (without the built in TypeScript enum type) or constants.

When a function offers a generic type parameter which extends string like T extends string, you can use that to enforce strong typing for your enums/constants.

function doSomething<T extends string>(t: T): {a: T} {
...
}

// invoke it
doSomething("foo" as const) // return {a: "foo"}

Don't make the mistake to lump extends in TypeScript's typesystem together with extends from ES classes - they are two complete distinct operators.

类extends对应instanceof于运行时并作为构造存在于运行时，而在类型系统中（extends例如，与条件类型一起发生），可以转换为“可分配给”，并且仅用于编译时。

更新：

您可以在TypeScript文档中找到有关字符串文字类型的更多信息。