Question

是否可以使用接受单个开关值的函数并返回布尔值来测试单个case语句？

如果没有，除了使用if语句之外，是否有更高效（速度或可读性）的方式来运行类似的测试？

下面举例说明。

在下面的代码中： IsInt（value）是一个函数，如果值是整数，则返回true或false isMultipleOfThree（value）是一个函数，如果值是三的倍数，则返回true或false。

var switchVal = "9";

switch( switchVal ){
  case(isMultipleOfThree):
    console.log(switchVal + ' is a multiple of three.');

  case(isInt):
    console.log(switchVal + ' is an integer.');
    break;

  default:
    console.log(switchVal + ' is not an integer or a multiple of three.');
 }

Answer 1

if else if怎么样？

var switchVal = "9";

if(isInt(switchVal)){
    console.log(switchVal + ' is an integer.');
}else if(isMultipleOfThree(switchVal)){
    console.log(switchVal + ' is a multiple of three.');
}else{
    console.log(switchVal + ' is not an integer or a multiple of three.');
}

然而，这并没有解决问题，如果它是9，它将退出并且只打印9是整数。您可以将它们分成不同的块。

var switchVal = "9";

if(isInt(switchVal)){
    console.log(switchVal + ' is an integer.');
if(isMultipleOfThree(switchVal)){
    console.log(switchVal + ' is a multiple of three.');
if(!isInt(switchVal) && !isMultipleOfThree(switchVal))
    console.log(switchVal + ' is not an integer or a multiple of three.');
}

但是当你添加越来越多的测试时，这可能会变得混乱。这当然假设您有两个函数isInt和isMultipleOfThree返回true或false。

采用带有标志的C方法的另一种奇怪的选择。

function isInt(s){
    if (!isNaN(s))
    return 1; //0001
  return 0;
}

function isMultipleOfThree(s){
    var int = parseInt(s);
    if(int % 3 === 0)
    return 2; //0010
  return 0;
}

function isSevem(s){
 var int = parseInt(s);
    if(int === 7)
    return 4; // 0100
  return 0;
}

var switchVal = "7";
console.log(7 % 3 === 0);
var result = isSevem(switchVal) + isMultipleOfThree(switchVal) + isInt(switchVal);
switch( result ){
  case 1: // 0001
    console.log(switchVal + ' is an integer.');
        break;
  case 2: // 0010
    console.log(switchVal + ' is a multiple of three.');
    break;
  case 3: // 0011
    console.log(switchVal + ' is an integer and a multiple of three.');
    break;
  case 4: // 0100
    console.log(switchVale + ' is the number seven.');
    break;
  case 5: // 0101
     console.log(switchVal + ' is an integer and is the number seven.');
     break;
  case 6: // 0110
    console.log(switchVal + ' is a multiple of three and is the number seven.');
    break;
  case 7: // 0111
    console.log(switchVal + ' is an integer and is a multiple of three and is the number seven.');
    break;

  default:
    console.log(switchVal + ' is not an integer or a multiple of three.');
    break;
 }

Answer 2

不，但你可以使用switch(true) {}做类似的事情：

var switchVal = "9";

switch( true ){
  case isInt(switchVal):
    console.log(switchVal + ' is an integer.');

  case isMultipleOfThree(switchVal):
    console.log(switchVal + ' is a multiple of three.');
    break;

  default:
    console.log(switchVal + ' is not an integer or a multiple of three.');
 }

如果您尝试执行所有传递案例，您可以编写一个执行此操作的函数：

function last(arr) {
  return arr[arr.length - 1];
}

function allCases(value, ...cases) {
  let defaultCase;
  let hasPassedAny = false;
  if(typeof last(cases) === "function") {
    defaultCase = last(cases);
    cases = cases.slice(0, -1);
  }
  cases.forEach(([predicate, resultFn]) => {
    if(predicate(value)) {
      hasPassedAny = true;
      resultFn(value);
    }
  });
  if(defaultCase && !hasPassedAny) {
    defaultCase(value);
  }
}

allCases(switchVal,
  [isInt, (v) => {
    console.log(`${v} is an integer`);
  }],
  [isMultipleOfThree, (v) => {
    console.log(`${v} is a multiple of three`);
  }],
  (v) => {
    console.log(`${v} is not an integer or a multiple of three.`);
  }
);

Javascript switch-case：使用单独的方法来评估case语句

2 个答案: