简单递归Javascript函数返回未定义

Question

对于赋值，我应该编写一个递归函数，使用N-2检查偶数或奇数的任何整数。如果even返回true，则返回false。但是只要值足够大就可以调用自身，它就会返回undefined。请帮忙！

function isEven(num) {
  console.log("top of function num = " + num);// For Debugging
  if (num == 0){
      return true;
  }
  else if (num == 1){
      return false;
  }
  else {
    num -= 2;
    console.log("num = " + num);
    isEven(num);
  }
}
console.log(isEven(0));
// → true
console.log(isEven(1));
// → false
console.log(isEven(8));
// → ??

控制台日志结果：

top of function num = 0

true

top of function num = 1

false

top of function num = 8

num = 6

top of function num = 6

num = 4

top of function num = 4

num = 2

top of function num = 2

num = 0

top of function num = 0

undefined

Answer 1

您在递归return来电之前忘记了isEven(num)声明。

请参阅下面的代码：

function isEven(num) {
  //console.log("top of function num = " + num);// For Debugging
  if (num == 0){
      return true;
  }
  else if (num == 1){
      return false;
  }
  else {
    num -= 2;
    return isEven(num);
  }
}
console.log('0 is even: ', isEven(0));
// → true
console.log('1 is even: ', isEven(1));
// → false
console.log('8 is even: ', isEven(8));

Answer 2

您可以更改以下行：

isEven(num);

到

return isEven(num);