我的数学函数返回 NaN

Question

所以我一直要求更改我的代码以简化我的功能，但现在我已经简化了它们，它们现在只返回 NaN 并且我真的不明白为什么会这样，当输入是数字时？

// the purpose of this script is to execute different functions on an array of 3 integers// 
const input = [6, 25, 11];
const findSum = input.reduce((acc, num) => acc + num, 0);
console.log(findSum);

// each function has been assigned parameters to execute //
const subsFrom = (num1, num2, num3) => num1 - num2 - num3;
const subNum = subsFrom(input);
console.log(subNum);
// then call the function and display the answer //

const multiBy = (num1, num2, num3) => num1 * num2 * num3;
const multiNum = multiBy(input)
console.log(multiNum);

const divBy = (findSum, num1, num2, num3) => findSum / num1 / num2 / num3;
const divNum = divBy(findSum, input);
console.log(divNum);

Answer 1

将数组解构为参数时，需要在参数列表中添加方括号。

const subsFrom = ([num1, num2, num3])
const multiBy = ([num1, num2, num3])
const divBy = (findSum, [num1, num2, num3])

因为输入是一个被传递到函数调用中的数组。

// the purpose of this script is to execute different functions on an array of 3 integers// 
const input = [6, 25, 11];
const findSum = input.reduce((acc, num) => acc + num, 0);
console.log(findSum);

// each function has been assigned parameters to execute //
const subsFrom = ([num1, num2, num3]) => num1 - num2 - num3;
const subNum = subsFrom(input);
console.log(subNum);
// then call the function and display the answer //

const multiBy = ([num1, num2, num3]) => num1 * num2 * num3;
const multiNum = multiBy(input)
console.log(multiNum);

const divBy = (findSum, [num1, num2, num3]) => findSum / num1 / num2 / num3;
const divNum = divBy(findSum, input);
console.log(divNum);

或者，您可以传播：

// the purpose of this script is to execute different functions on an array of 3 integers// 
const input = [6, 25, 11];
const findSum = input.reduce((acc, num) => acc + num, 0);
console.log(findSum);

// each function has been assigned parameters to execute //
const subsFrom = (num1, num2, num3) => num1 - num2 - num3;
const subNum = subsFrom(...input);
console.log(subNum);
// then call the function and display the answer //

const multiBy = (num1, num2, num3) => num1 * num2 * num3;
const multiNum = multiBy(...input)
console.log(multiNum);

const divBy = (findSum, num1, num2, num3) => findSum / num1 / num2 / num3;
const divNum = divBy(findSum, ...input);
console.log(divNum);