javascript - 为什么 switch case 中的逻辑 OR 运算符会表现得很奇怪？

Question

testMyNumber函数似乎认为第二个位置的数字不匹配。

为什么 4 和 10 返回默认情况？

function testMyNumber(number) {
  switch (number) {
    case 6:
      return number+" is 6";
      break;
    case (3 || 4) :
      return number+" is 3 or 4";
      break;

    case 9 || 10 :
      return number+" is 9 or 10";
      break;

    default:
      return number+" is not in 3,4,6,9,10";
      break;
  }
};
console.log(testMyNumber(6));
console.log(testMyNumber(3));
console.log(testMyNumber(4));
console.log(testMyNumber(9));
console.log(testMyNumber(10));

有没有办法使这项工作？

Answer 1

因为它不打算以这种方式使用。||如果它是真实的，则返回它的第一个操作数，否则它的第二个操作数。3 || 4返回 3因为3是真实的，因此case只会检查 3 :

console.log(3 || 4); //3, because it is truthy
console.log(0 || 1); //1, because 0 is falsy

为了使您的代码正常工作，请使用单独的案例，这些案例包括:

function testMyNumber(number) {
  switch (number) {
    case 6:
      return number+" is 6";
    case 3:
    case 4:
      return number+" is 3 or 4";

    case 9:
    case 10:
      return number+" is 9 or 10";
    default:
      return number+" is not in 3,4,6,9,10";
  }
};
console.log(testMyNumber(6));
console.log(testMyNumber(3));
console.log(testMyNumber(4));
console.log(testMyNumber(9));
console.log(testMyNumber(10));