有人可以将此箭头函数转换为普通函数吗？

Question

我无法将此箭头函数转换为普通函数。我已经在 chrome 的控制台面板中对此进行了测试。此代码取自 freecodeCamp.org 在 Es6 课

//This is what I have tried. The final output result is showing undefined


const realNumberArray = [4, 5.6, -9.8, 3.14, 42, 6, 8.34, -2];
const squareList = function(arr) {
    "use strict";
    const squaredIntegers = function(num) {
        (function() {
            arr.filter(Number.isInteger(num) && num > 0);
        });       
        return squaredIntegers;
    } 
}

const squaredIntegers = squareList(realNumberArray);
console.log(squaredIntegers);

//Here is the Arrow function I was trying to convert

const realNumberArray = [4, 5.6, -9.8, 3.14, 42, 6, 8.34, -2];
const squareList = (arr) => {
    "use strict";
    const squaredIntegers = arr.filter(num => Number.isInteger(num) && num > 0);
    return squaredIntegers;
};
const squaredIntegers = squareList(realNumberArray);
console.log(squaredIntegers);


//The code should output this
[4, 42, 6];

Answer 1

我将你的代码插入 Babel 并得到了这个：

"use strict";

var realNumberArray = [4, 5.6, -9.8, 3.14, 42, 6, 8.34, -2];

var squareList = function squareList(arr) {
  "use strict";

  var squaredIntegers = arr.filter(function (num) {
    return Number.isInteger(num) && num > 0;
  });
  return squaredIntegers;
};

var squaredIntegers = squareList(realNumberArray);
console.log(squaredIntegers);

（一般情况下，当你需要将ES6+语法转为ES5时，可以使用Babel自动完成。）

Answer 2

箭头函数中 => 之后的任何表达式都成为函数的隐式 return，但您需要在普通函数中使用 return 关键字显式 return。

const realNumberArray = [4, 5.6, -9.8, 3.14, 42, 6, 8.34, -2];
const squareList = function(arr){
    "use strict";
    const squaredIntegers = arr.filter(function(num){
        return Number.isInteger(num) && num > 0
    });
    return squaredIntegers;
};
const squaredIntegers = squareList(realNumberArray);
console.log(squaredIntegers);

Answer 3

你那里有一些问题。

函数语法是

function name(param_list)
function name(p1,pn)

也许保留内联声明更好。

function natural(num)
{
   return Number.isInteger(num) && num > 0;
}

function squarelist(arr)
{
    "use strict";
    const squaredIntegers = arr.filter(natural); //notice non lambda needs a name
    return squaredIntegers;
}

Answer 4

var realNumberArray = [4, 5.6, -9.8, 3.14, 42, 6, 8.34, -2];

var squareList = function squareList(arr) {
  "use strict";

  var squaredIntegers = arr.filter(function (num) {
    return Number.isInteger(num) && num > 0;
  });
  return squaredIntegers;
};

var squaredIntegers = squareList(realNumberArray);
console.log(squaredIntegers);

我想这很成功。

有人可以将此箭头函数转换为普通函数吗？

Can someone convert this arrow function into Normal function?

javascript

web

ecmascript-6

arrow-functions