三元运算符不返回未定义

Question

我正在研究 FCC 中间算法 "Arguments Optional"。以下是有关需要发生的事情的说明：

Intermediate Algorithm Scripting: Arguments Optional

创建一个将两个参数相加的函数。如果只提供一个参数，那么 return 一个函数需要一个参数并且 return 是总和。
例如，addTogether(2, 3) 应该 return 5，而 addTogether(2) 应该 return 一个函数。
使用单个参数调用此 returned 函数将 return 求和： var sumTwoAnd = addTogether(2); sumTwoAnd(3) returns 5.
如果任一参数不是有效数字，return未定义。

我写了代码来完成上面解释的所有事情，但一个要求是参数必须全部是数字，否则 return undefined（上面的#4）。您会看到我写了一个三元运算符（代码的第 5 行）numbersOnly 变量，我认为它可以处理这个问题，但它只是在控制台中 returning [Function]。

function addTogether() {
    // Convert args to an array
    let args = [...arguments];
    // Only accept numbers or return undefined and stop the program
    const numbersOnly = value => typeof(value) === 'number'? value : undefined;
    // test args for numbersOnly and return only the first two arguments regardless of the length of args
    let numbers = args.filter(numbersOnly).slice(0, 2);

    // // It has to add two numbers passed as parameters and return the sum.
    if (numbers.length > 1) {
        return numbers[0] + numbers[1];
    }
    // If it has only one argument then it has to return a function that uses that number and expects another one, to then add it.
    else if (numbers.length === 1) {
        let firstParam = numbers[0];
        return function(secondParam) {
            if (typeof secondParam !== 'number' || typeof firstParam !== 'number') {
                return undefined;
            }
            return secondParam + firstParam;
        }
    }
}

我正在通过所有测试，除了 #4，它应该 return 未定义。我不太明白为什么 5 会通过而 returning 未定义但 4 会失败。我在这里错过了什么？谢谢！

1. addTogether(2, 3) should return 5.
2. addTogether(2)(3) should return 5.
3. addTogether("https://www.youtube.com/watch?v=dQw4w9WgXcQ") should return undefined.
4. addTogether(2, "3") should return undefined.
5. addTogether(2)([3]) should return undefined.

Answer 1

因为过滤参数时，只过滤掉'3'，所以剩下的数组长度为1。

const args = [2, '3'];
const numbers = args.filter((n) => typeof n === 'number');
console.log('numbers:', numbers);
console.log('numbers.length === 1:', numbers.length === 1);

Answer 2

这是因为您必须检查过滤器的输入参数和输出参数。尝试添加此代码段：

let numbers = args.filter(numbersOnly).slice(0, 2);
if (args.length > numbers.length) {
  return undefined;
}

function addTogether() {
    // Convert args to an array
    let args = [...arguments];
    // Only accept numbers or return undefined and stop the program
    const numbersOnly = value => typeof(value) === 'number'? value : undefined;
    // test args for numbersOnly and return only the first two arguments regardless of the length of args
    let numbers = args.filter(numbersOnly).slice(0, 2);
    if (args.length > numbers.length) {
      return undefined;
    }

    // // It has to add two numbers passed as parameters and return the sum.
    if (numbers.length > 1) {
        return numbers[0] + numbers[1];
    }
    // If it has only one argument then it has to return a function that uses that number and expects another one, to then add it.
    else if (numbers.length === 1) {
        let firstParam = numbers[0];
        return function(secondParam) {
            if (typeof secondParam !== 'number' || typeof firstParam !== 'number') {
                return undefined;
            }
            return secondParam + firstParam;
        }
    }
  }
  console.log('4. addTogether', addTogether(4, "4"));

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