TypeScript-函数兼容性

参数个数

• 可以将参数少的函数赋值给参数多的函数

let fn1 = (x: number, y: number) => {
};
let fn2 = (x: number) => {
};

fn1 = fn2;

• ​​不可以​​ 将参数多的函数赋值给参数少的函数

let fn1 = (x: number, y: number) => {
};
let fn2 = (x: number) => {
};

fn2 = fn1;

参数类型

• ​​可以​​ 将参数类型一样的函数赋值给参数类型一样的函数

let fn1 = (x: number) => {
};
let fn2 = (x: number) => {
};
let fn3 = (x: string) => {
};
fn1 = fn2;
fn2 = fn1;

• ​​不可以​​ 将参数类型不一样的函数赋值给参数类型不一样的函数, 必须一模一样

let fn1 = (x: number) => {
};
let fn2 = (x: number) => {
};
let fn3 = (x: string) => {
};

fn1 = fn3;
fn3 = fn1;

返回值类型

• ​​可以​​ 将返回值类型一样的函数赋值给返回值类型一样的函数

let fn1 = (): number => 123;
let fn2 = (): number => 456;
let fn3 = (): string => 'abc';
fn1 = fn2;
fn2 = fn1;

• ​​不可以​​ 将返回值类型不一样的函数赋值给返回值类型不一样的函数，必须一模一样

let fn1 = (): number => 123;
let fn2 = (): number => 456;
let fn3 = (): string => 'abc';

fn1 = fn3;
fn3 = fn1;

函数双向协变

参数的双向协变

let fn1 = (x: (number | string)) => {
};
let fn2 = (x: number) => {
};
fn1 = fn2;
fn2 = fn1;

返回值双向协变

• 但是可以将返回值是具体类型的赋值给联合类型的

let fn1 = (x: boolean): (number | string) => x ? 123 : 'abc';
let fn2 = (x: boolean): number => 456;
fn1 = fn2;

• 不能将返回值是联合类型的赋值给具体类型的

let fn1 = (x: boolean): (number | string) => x ? 123 : 'abc';
let fn2 = (x: boolean): number => 456;
fn2 = fn1;

函数重载

function add(x: number, y: number): number;
function add(x: string, y: string): string;
function add(x, y) {
    return x + y;
}

function sub(x: number, y: number): number;
function sub(x, y) {
    return x - y;
}

• 不能将重载少的赋值给重载多的

function add(x: number, y: number): number;
function add(x: string, y: string): string;
function add(x, y) {
    return x + y;
}

function sub(x: number, y: number): number;
function sub(x, y) {
    return x - y;
}

let fn = add;
fn = sub;

• 可以将重载多的赋值给重载少

function add(x: number, y: number): number;
function add(x: string, y: string): string;
function add(x, y) {
    return x + y;
}

function sub(x: number, y: number): number;
function sub(x, y) {
    return x - y;
}

let fn = sub;
fn = add;