Javascript函数transfomr我的未定义值。(Javascript function transfomr my value in undefined. Me and my teacher

我今天在这里，因为我有一个错误我（和我的老师）在javascript中不理解。

以下代码只是函数的一个因子（强制用函数执行）。 但是返回时我的值在undefined中被改变了。

var test;
do {
  var result = parseInt(prompt("Un chiffre supérieur à 1"));
} while (isNaN(result) || result < 2);
// Just asking for a number > à 1 it work //

test = facto(result, 1);
console.log(test);
// The test variable was used to try to understand the bug //

function facto(chiffre, fact) {
  // Function for the factorial //

  fact = fact * chiffre;
  // fact was defined as 1 for the first entry, for others it's gonna be his last  result //`
  console.log(fact);
  chiffre = chiffre - 1;
  // We decrase chiffre by one for the next passage //


  if (chiffre == 1) {
    // if chiffre = 1, we don't need to do other passage, so we return value 
    console.log(fact);
    return fact;
    // the result is correct, it's a number. However it return undefined //
  } else {
    // if chiffre > 1, we continue the math by coming again in the function //
    console.log(chiffre);
    facto(chiffre, fact);
  }
} 
  
 
 
 我不明白为什么它返回undefined。 有人可以解释一下吗？ 此外，如果函数只执行一次，该函数也可以工作。 
 
 先谢谢你。 
I'm here today because i've an error I (and my teacher) don't understand in javascript.
 
The following code is just an factorial by a function (forced to do it with function). But my value is changed in undefined when return.
 
 
 
  
  
var test;
do {
  var result = parseInt(prompt("Un chiffre supérieur à 1"));
} while (isNaN(result) || result < 2);
// Just asking for a number > à 1 it work //

test = facto(result, 1);
console.log(test);
// The test variable was used to try to understand the bug //

function facto(chiffre, fact) {
  // Function for the factorial //

  fact = fact * chiffre;
  // fact was defined as 1 for the first entry, for others it's gonna be his last  result //`
  console.log(fact);
  chiffre = chiffre - 1;
  // We decrase chiffre by one for the next passage //


  if (chiffre == 1) {
    // if chiffre = 1, we don't need to do other passage, so we return value 
    console.log(fact);
    return fact;
    // the result is correct, it's a number. However it return undefined //
  } else {
    // if chiffre > 1, we continue the math by coming again in the function //
    console.log(chiffre);
    facto(chiffre, fact);
  }
} 
  
 
 
I don't understand why it return undefined. Could someone explain me ? Also, the function work if it go only one time in the function.
 
Thank you in advance.
                
最满意答案
                
                    
                        
 你必须返回递归调用，否则第一次调用facto将永远不会遇到return语句，导致第一次调用（默认情况下）返回undefined 。 改为return facto(chiffre, fact); ： 
 
 
 
  
  
var test;
do {
  var result = parseInt(prompt("Un chiffre supérieur à 1"));
} while (isNaN(result) || result < 2);
test = facto(result, 1);
console.log(test);

function facto(chiffre, fact) {
  fact = fact * chiffre;
  chiffre = chiffre - 1;
  if (chiffre == 1) {
    return fact;
  } else {
    return facto(chiffre, fact);
  }
} 
  
 
You have to return the recursive call, else the first call of facto will never encounter a return statement, resulting in that first call (by default) returning undefined. Change to return facto(chiffre, fact);:
 
 
 
  
  
var test;
do {
  var result = parseInt(prompt("Un chiffre supérieur à 1"));
} while (isNaN(result) || result < 2);
test = facto(result, 1);
console.log(test);

function facto(chiffre, fact) {
  fact = fact * chiffre;
  chiffre = chiffre - 1;
  if (chiffre == 1) {
    return fact;
  } else {
    return facto(chiffre, fact);
  }
}