为什么这个阶乘函数不返回？

Question

我创建了这个递归函数来计算一个数字的阶乘。第一个参数 n 是您要为其计算阶乘的数字，第二个参数 result 用于在调用自身时将阶乘计算的状态传递给函数。我遇到的问题是该函数将在函数结束时控制台记录正确的阶乘结果，但不会返回它。它只会返回小于 2 的数字的阶乘，所有其他返回未定义。我有一个带有“n > = 2”的条件，所以这让我认为这与此有关，但是我找不到这与问题之间的任何关系。为什么这个函数没有返回正确的阶乘？

function factorial(n, result){

    //checks if result is undefined if so uses n calculate value of result
    //if result isnt undefined it changes its value using "n * ( n - 1)"

    result = result * (n - 1) || n * (n - 1);



   //decreases n, n gets gradually smaller
    n-=1;


 //if n is more the 2 run function again
 //I'm fairly certain this conditional is the root of the issue but personally cant find the relation
    if(n >= 2){
        //passes current state of n and result
        factorial(n,result);
       }
    else {
       //result has the correct value as its printing to the console 
       //correctly, e.g !4 = 4*3*2*1, which equals 24, this prints 24 if you 
       //pass 4 to n.
       console.log(result);
       //if n is smaller then 2 return factorial
       //but it wont return result
        return result;
        };
}

Answer 1

你的函数没有返回结果的原因是你在n >=2时没有返回factorial(n, result)的结果

function factorial(n, result){

    //checks if result is undefined if so uses n calculate value of result
    //if result isnt undefined it changes its value using "n * ( n - 1)"

    result = result * (n - 1) || n * (n - 1);



   //decreases n, n gets gradually smaller
    n-=1;


 //if n is more the 2 run function again
 //I'm fairly certain this conditional is the root of the issue but personally cant find the relation
    if(n >= 2){
        //passes current state of n and result
        return factorial(n,result);
       }
    else {
       //result has the correct value as its printing to the console 
       //correctly, e.g !4 = 4*3*2*1, which equals 24, this prints 24 if you 
       //pass 4 to n.
       console.log(result);
       //if n is smaller then 2 return factorial
       //but it wont return result
        return result;
        };
}

但是，编写此代码的最简单方法是递归直到达到终止条件

function factorial(n) {
   if(n == 1) {
       return 1;
   }
   return n*factorial(n - 1);
}