TypeScript中的重载函数类型

我们定义一个简单的重载函数:

typescript

function fn(): void;
function fn(a: string): string;
function fn(a: number): number;
function fn(a: string, b: number): [string, number];
function fn(a?: string | number, b?: number): string | number | [string, number] | void {
  if (a !== undefined && b !== undefined) {
    return [a.toString(), b];
  }
  if (a !== undefined) {
    return a;
  }
  return;
}

然后尝试获取函数的参数类型:

typescript

type ParametersFn = Parameters<typeof fn>
type ReturnTypeFn = ReturnType<typeof fn>

此时我们发现, 只有最后一个overload签名的参数类型和返回值类型被返回了...

那我们到底如何才能正确获取重载函数的所有参数类型和返回值类型呢?

首先, 我们来研究一下如何正确的定义一个重载函数类型, 参考TypeScript handbook中的Call Signatures, 我们很容易写出上面的重载函数的类型:

typescript

interface OverloadFn {
  (): void;
  (a: string): string;
  (a: number): number;
  (a: string, b: number): [string, number];
}

接着, 试着写出一个OverloadReturnType<T>来获取返回值类型, 但是问题在于overload签名的数量不是固定的, 如何来进行定义呢? 这里只能采用一个折中的写法, 以下写法最多可以支持8个函数重载签名:

typescript

type OverloadedReturnType<T> = 
  T extends { (...args: any[]): infer R; (...args: any[]): infer R; (...args: any[]): infer R; (...args: any[]): infer R; (...args: any[]): infer R; (...args: any[]): infer R; (...args: any[]): infer R; (...args: any[]): infer R; } ? R  :
  T extends { (...args: any[]): infer R; (...args: any[]): infer R; (...args: any[]): infer R; (...args: any[]): infer R; (...args: any[]): infer R; (...args: any[]): infer R; (...args: any[]): infer R; } ? R :
  T extends { (...args: any[]): infer R; (...args: any[]): infer R; (...args: any[]): infer R; (...args: any[]): infer R; (...args: any[]): infer R; (...args: any[]): infer R; } ? R :
  T extends { (...args: any[]): infer R; (...args: any[]): infer R; (...args: any[]): infer R; (...args: any[]): infer R; (...args: any[]): infer R; } ? R :
  T extends { (...args: any[]): infer R; (...args: any[]): infer R; (...args: any[]): infer R; (...args: any[]): infer R } ? R :
  T extends { (...args: any[]): infer R; (...args: any[]): infer R; (...args: any[]): infer R; } ? R  :
  T extends { (...args: any[]): infer R; (...args: any[]): infer R; } ? R :
  T extends { (...args: any[]): infer R; } ? R : any;

我们尝试着使用一下, 结果非常完美!

接下来, 大家信心满满的写出了OverloadParameters<T>, 想着今天终于可以不用加班了, 然而 ...

typescript

type OverloadedParameters<T> = 
  T extends { (...args: infer P) : any; (...args: infer P) : any; (...args: infer P) : any ; (...args: infer P) : any; (...args: infer P) : any; (...args: infer P) : any; (...args: infer P) : any; (...args: infer P) : any } ? P :
  T extends { (...args: infer P) : any; (...args: infer P) : any; (...args: infer P) : any ; (...args: infer P) : any; (...args: infer P) : any; (...args: infer P) : any; (...args: infer P) : any } ? P :
  T extends { (...args: infer P) : any; (...args: infer P) : any; (...args: infer P) : any ; (...args: infer P) : any; (...args: infer P) : any; (...args: infer P) : any } ? P :
  T extends { (...args: infer P) : any; (...args: infer P) : any; (...args: infer P) : any ; (...args: infer P) : any; (...args: infer P) : any } ? P :
  T extends { (...args: infer P) : any; (...args: infer P) : any; (...args: infer P) : any ; (...args: infer P) : any } ? P :
  T extends { (...args: infer P) : any; (...args: infer P) : any; (...args: infer P) : any } ? P :
  T extends { (...args: infer P) : any; (...args: infer P) : any } ? P :
  T extends { (...args: infer P) : any; } ? P : any;

type OverloadParameters = OverloadedParameters<typeof fn>;

查阅TypeScrpit相关源码和Issues后发现, 条件类型中infer关键字后面推导的类型, 通常是做Union连接, 例如return type, parameter type, type parameter以及union type中出现的类型:

typescript

type ItemType<T> = T extends Array<infer I> ? I : T;
type Item = ItemType<string | number[] | boolean[]>;

type ReturnTypes<T> = T extends { get(): infer G; pop(): infer G; } ? G : never;
type Return = ReturnTypes<{ get(): string; pop(): number}>;

但infer右值类型是重载函数的parameter type时, 却是采用intersection type, 所以OverloadedParameters<T>得到的结果是void & string & number & [string, number]

typescript

type ParameterTypes<T> = T extends { set(a: infer P): any; push(b: infer P): ; } ? P : any;
type Params = ParameterTypes<{ set(a: string): any; push(b: number): any}>;

毫无疑问, 交集类型的结果是never, 因为不存在一个类型同时是void string number 和 [string, number]的子集 (参考NonNullable源码)

具体情况参考:

https://github.com/microsoft/TypeScript/blob/v5.0.4/src/deprecatedCompat/deprecations.ts#L65
https://github.com/microsoft/TypeScript/blob/v5.0.4/src/deprecatedCompat/deprecations.ts#L82
https://github.com/microsoft/TypeScript/blob/v5.0.4/src/compiler/types.ts#L9781
https://github.com/microsoft/TypeScript/issues/14107
https://github.com/microsoft/TypeScript/issues/32164

所以首先要把重载函数类型转换成其他类型, 例如函数元组(Tuple), 以下代码可以支持最多8次重载签名;

typescript

type OverloadsToTuple8<T> = T extends { (...args: infer P1): infer R1; (...args: infer P2): infer R2; (...args: infer P3): infer R3; (...args: infer P4): infer R4; (...args: infer P5): infer R5; (...args: infer P6): infer R6; (...args: infer P7): infer R7; (...args: infer P8): infer R8; } 
    ? [ (...args: P1) => R1, (...args: P2) => R2, (...args: P3) => R3, (...args: P4) => R4, (...args: P5) => R5, (...args: P6) => R6, (...args: P7) => R7, (...args: P8) => R8 ] 
    : OverloadsToTuple7<T>;

type OverloadsToTuple7<T> = T extends { (...args: infer P1): infer R1; (...args: infer P2): infer R2; (...args: infer P3): infer R3; (...args: infer P4): infer R4; (...args: infer P5): infer R5; (...args: infer P6): infer R6; (...args: infer P7): infer R7; } 
    ? [ (...args: P1) => R1, (...args: P2) => R2, (...args: P3) => R3, (...args: P4) => R4, (...args: P5) => R5, (...args: P6) => R6, (...args: P7) => R7 ] 
    : OverloadsToTuple6<T>;

type OverloadsToTuple6<T> = T extends { (...args: infer P1): infer R1; (...args: infer P2): infer R2; (...args: infer P3): infer R3; (...args: infer P4): infer R4; (...args: infer P5): infer R5; (...args: infer P6): infer R6; }
    ? [ (...args: P1) => R1, (...args: P2) => R2, (...args: P3) => R3, (...args: P4) => R4, (...args: P5) => R5, (...args: P6) => R6 ] 
    : OverloadsToTuple5<T>;

type OverloadsToTuple5<T> = T extends { (...args: infer P1): infer R1; (...args: infer P2): infer R2; (...args: infer P3): infer R3; (...args: infer P4): infer R4; (...args: infer P5): infer R5; }
    ? [ (...args: P1) => R1, (...args: P2) => R2, (...args: P3) => R3, (...args: P4) => R4, (...args: P5) => R5 ] 
    : OverloadsToTuple4<T>;

type OverloadsToTuple4<T> = T extends { (...args: infer P1): infer R1; (...args: infer P2): infer R2; (...args: infer P3): infer R3; (...args: infer P4): infer R4; } 
    ? [ (...args: P1) => R1, (...args: P2) => R2, (...args: P3) => R3, (...args: P4) => R4 ] : OverloadsToTuple3<T>;

type OverloadsToTuple3<T> = T extends { (...args: infer P1): infer R1; (...args: infer P2): infer R2; (...args: infer P3): infer R3; }
    ? [ (...args: P1) => R1, (...args: P2) => R2, (...args: P3) => R3, ] : OverloadsToTuple2<T>;

type OverloadsToTuple2<T> = T extends { (...args: infer P1): infer R1; (...args: infer P2): infer R2; } 
    ? [ (...args: P1) => R1, (...args: P2) => R2 ]
    : OverloadsToTuple1<T>;

type OverloadsToTuple1<T> = T extends { (...args: infer P1): infer R1; } ? [ (...args: P1) => R1 ] : never;

type OverloadsToTuple<T> = OverloadsToTuple8<T>;

简单测试后我们发现, 重载次数小于8时, 会出现重复的(...args: unknown[]) => unknown签名:

所以, 尝试过滤多余的元组项:

typescript

type FilterUnknowSignature<T extends unknown[]> = T extends [] ? [] :
    T extends [infer H, ...infer R] 
        ? ((...args: unknown[]) => unknown) extends H
            ? FilterUnknowSignature<R> 
            : [H, ...FilterUnknowSignature<R>] 
        : T

type OverloadsToTuple<T> = FilterUnknowSignature<OverloadsToTuple8<T>>;

type Tuple = OverloadsToTuple<type fn>;

但是发现() => void签名也被过滤了....

TypeScript中的重载函数类型 ​

TypeScript中的重载函数类型