在JavaScript中尋找switch陳述句的干凈函式實作

我正在教授一門課程，其中包括解釋函式式 JavaScript，我希望有一個非常好的函式式編程示例，希望它比非函式式更簡潔。我想將以下 switch 陳述句轉換為函式式。我自己做了一個轉換的例子，但希望有一個更簡單的解決方案。

這是 switch 陳述句版本：

let animalType = "Poodle";
switch (animalType) {
  case "Poodle":
  case "Beagle":
  case "Bulldog":
    console.log(animalType   " is a dog.");
    break;
  case "Bengal":
  case "Siamese":
    console.log(animalType   " is a cat.");
    break;
  default:
    console.log(animalType   " is not a dog or cat.");
    break;
}

這就是我想出的功能，但我對此并不滿意

const result = getAnimalType("Poodle");
console.log("result:"   result)

function getAnimalType(animal) {
  function isDog(animal) {
    const dogs = ["Poodle", "Beagle", "Bulldog"];
    return dogs.includes(animal)
  }
  function isCat(animal) {
    const cats = ["Bengal", "Siamese"];
    return cats.includes(animal)
  }
  return isDog(animal)
    ? animal   " is a dog."
    : isCat(animal)
    ? animal   " is a cat."
    : animal   " is not a dog or cat.";
}

uj5u.com熱心網友回復：

一種選擇是由 dog 或 cat 索引的物件，其值是動物型別的陣列。這很容易擴展到其他動物型別。

const animalNamesByType = {
  dog: ["Poodle", "Beagle", "Bulldog"],
  cat: ["Bengal", "Siamese"]
};
function getAnimalType(animal) {
  const entry = Object.entries(animalNamesByType).find(
    entry => entry[1].includes(animal)
  );
  return entry
  ? `${animal} is a ${entry[0]}`
  : `${animal} is not in animalNamesByType`;
}
console.log(getAnimalType("Poodle"));

uj5u.com熱心網友回復：

您可以使用物件將動物型別映射到函式。

function dog(animalType) {
  return animalType   " is a dog.";
}

function cat(animalType) {
  return animalType   " is a cat.";
}

function other(animalType) {
  return animalType   " is not a dog or cat.";
}

const typeMap = {
  Poodle: dog,
  Beagle: dog,
  Bulldog: dog,
  Bengal: cat,
  Siamese: cat
};

function getAnimalType(animalType) {
  let typeFun = typeMap[animalType] || other;
  return typeFun(animalType);
}

console.log(getAnimalType("Poodle"));

uj5u.com熱心網友回復：

您可以為此創建一個非常簡單的 3 行函式

const dogs = ["Poodle", "Beagle", "Bulldog"];
const cats = ["Bengal", "Siamese"];

const getAnimalType = (animal) => {
  if(dogs.includes(animal)) return `${animal} is a dog`
  if(cats.includes(animal)) return `${animal} is a cat`
  return `${animal} is not a dog or cat.`
}

const result = getAnimalType("Poodle");
console.log("result:"   result)

uj5u.com熱心網友回復：

一個變種，有點像 Barmar 的，但對我來說仍然是個人的

const getAnimalType = (() => 
  {
  const
    isX =
    { dog : 'is a dog'
    , cat : 'is a cat'
    , nDC : 'is neither a dog nor a cat'
    }
  , typeMap = 
    { Poodle  : 'dog'
    , Beagle  : 'dog'
    , Bulldog : 'dog'
    , Bengal  : 'cat'
    , Siamese : 'cat'
    };
  return (animal) => `${animal} ${isX[ typeMap[animal] ?? 'nDC']}`
  })()

console.log(getAnimalType('Beagle'))
console.log(getAnimalType('Bengal'))
console.log(getAnimalType('schtroumpf'))

.as-console-wrapper {max-height: 100% !important;top: 0;}
.as-console-row::after {display: none !important;}

uj5u.com熱心網友回復：

switch從函式式編程的角度來看并沒有錯。問題在于它console.log作為副作用呼叫，而不是回傳值。不過很容易修復：

function getAnimalType(animalType) {
  switch (animalType) {
    case "Poodle":
    case "Beagle":
    case "Bulldog":
      return animalType   " is a dog.";
    case "Bengal":
    case "Siamese":
      return animalType   " is a cat.";
    default:
      return animalType   " is not a dog or cat.";
  }
}

console.log(getAnimalType("Poodle"));

下一個改進可能是避免一些重復：

function getAnimalType(animalType) {
  switch (animalType) {
    case "Poodle":
    case "Beagle":
    case "Bulldog":
      return "a dog";
    case "Bengal":
    case "Siamese":
      return "a cat";
    default:
      return "not a dog or cat";
  }
}
function getStatement(animalType) {
  return animalType   " is "   getAnimalType(animalType)   ".";
}
console.log(getStatement("Poodle"));

uj5u.com熱心網友回復：

通過編譯 TypeScript 列舉代碼示例，我得到了與此類似的東西：

var AnimalType;
(function (AnimalType) {
    AnimalType[AnimalType["Poodle"] = 0] = "Poodle";
    AnimalType[AnimalType["Beagle"] = 1] = "Beagle";
    AnimalType[AnimalType["Bulldog"] = 2] = "Bulldog";
    AnimalType[AnimalType["Bengal"] = 3] = "Bengal";
    AnimalType[AnimalType["Siamese"] = 4] = "Siamese";
})(AnimalType || (AnimalType = {}));

function whichKind(animalType) {
    let kind;
    switch (animalType) {
        case AnimalType.Poodle:
            kind = `${AnimalType[animalType]} is a dog`;
            break;
        case AnimalType.Beagle:
            kind = `${AnimalType[animalType]} is a dog`;
            break;
        case AnimalType.Bulldog:
            kind= `${AnimalType[animalType]} is a dog`;
            break;
       case AnimalType.Bengal:
            kind = `${AnimalType[animalType]} is a cat`;
            break;
       case AnimalType.Siamese:
            kind = `${AnimalType[animalType]} is a cat`;
           break;
    }
 
    kind = typeof kind === 'undefined' ? `${animalType} is not a dog or cat.` : kind;

    return kind;
}

console.log(whichKind(AnimalType.Poodle)); // Poodle is a dog

// console.log(whichKind('Other')); // Other is not a dog or cat.

標籤：javascript 函数式编程

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