(1) infer 出现在extends 条件语句后的函数类型的参数位置上
interface Customer {
custname:string
buymoney:number
}
type custFuncType = (cust:Customer) => string
type inferType<T> = T extends (params:infer P)=> any?P:T
type inferResultType = inferType<custFuncType>
解析 type inferType<(cust:Customer) => string> = (cust:Customer) => string extends (params:infer P)=> any?P:T (cust:Customer) => string 和 (params:infer P)=> any 匹配 P替换为Customer 结果为true 返回 Customer
(2) infer 出现在extends 条件语句后的函数类型的返回值类型上
interface Customer {
custname:string
buymoney:number
}
type custFuncType = (cust:Customer) => string
type inferType<T> = T extends (params:any)=> infer P ? P:T
type inferResultType = inferType<custFuncType>
解析 type inferType<(cust:Customer) => string> = (cust:Customer) => string extends (params:any)=> infer P ? P:T (cust:Customer) => string 和 (params:any)=> infer P 匹配 P替换为string 结果为true 返回 string
(3) infer 出现在类型的泛型具体化类型上
type Elementofo<T> = T extends Set<infer E> ? E : never
let result:Elementofo<Set<string>>
解析 Elementofo<Set> ===>> Set extends Set E 得到 string 类型 返回结果 string
infer的使用场景
class TestClass {
public name:string
public classno:number
constructor(name:string,classno:number){
this.name = name
this.classno = classno
}
eat(){
console.log("姓名为："+this.name+" 班级:"+this.classno)
}
}
type ConstructorParametersType<T> = T extends new (...args: infer P) => any ? P : never
type Constructor<T> = new (...args:any[]) => T
//let constructor:ConstructorParametersType<typeof TestClass> //[name: string, classno: number]
function createInstance<T,D>(constructor:Constructor<T>,...args:ConstructorParametersType<D>){
return new constructor(args[0],args[1])
}
createInstance<TestClass,typeof TestClass>(TestClass,"张三",18).eat()
