如何期望函数的两个参数具有相同的特定 Union 类型值

Question

我试图创建一个带有两个参数的函数，第一个是联合类型，第二个是取决于第一个参数类型的类型。

例如：

type Fruit = "Orange" | "Apple" | "Banana";

function doubleFruit<T extends Fruit, K extends T>(fruit1: T, fruit2: K) {
   // When execution reaches here
   // both fruit1 and fruit2 need to be the same fruit
}

---

现在，这些语句按预期工作

 
 // Gives error since Orange is not same as Apple neither banana
 doubleFruit("Orange", "Apple"); 
 doubleFruit("Orange", "Banana");

 // Runs peacefully, since both are same fruit
 doubleFruit("Apple", "Apple");

---

但是当第一个参数是动态的时候，奇怪的事情发生了

function run(x: "Orange" | "Apple") {

 // Should not work, since x could be anything
 doubleFruit(x, "Orange");
}

我假设因为x可以是Orange或Apple，所以第二个参数Orange不遵循两个参数相同的标准。

编辑：这就是我想要实现的目标

type Fruit = "Orange" | "Apple" | "Banana";


function doubleFruit<T extends Fruit, K extends T>(fruit1: T, fruit2: K) {
  // When execution reaches here
  // both fruit1 and fruit2 need to be the same fruit
}

function run(x: "Orange" | "Apple") {
  // I need this to work: 
  //   Whatever be the value of x, the other parameter would be equivalent,
  //   hence should be allowed
  doubleFruit(x, x); 

  // But not these: 
  //   since x could be anything, specifically sending one of 
  //   values shouldn't work
  doubleFruit(x, "Orange")
  doubleFruit(x, "Apple")
  
}

我需要一种方法来满足上述行为。

Answer 1

请尝试这是否适合您的情况：

type Fruit = "Orange" | "Apple" | "Banana";

function doubleFruit<T extends Fruit, K>(fruit1: T, fruit2: K & T) {
   // When execution reaches here
   // both fruit1 and fruit2 need to be the same fruit
}

doubleFruit("Orange", "Apple"); 
doubleFruit("Orange", "Banana");

doubleFruit("Apple", "Apple");

const x: "Orange" | "Apple" = "Apple";

doubleFruit(x, "Orange");