javascript - 累加器在 Javascript 的 reduce 函数中返回 NaN

Question

我正在学习使用扩展运算符...来编写一个函数，该函数接受传递给函数的所有参数并返回偶数的总和。我的问题是，为什么除了reduce()的第一个回调之外，我的acc等于NaN？

代码和执行的打印输出如下，console.log(...)是我插入的调试代码。感谢您的帮助。

function sumEvenArgs(...args){
      var sum = args.reduce( (acc, next) => {
        console.log("\nnext:", next);
        if (next % 2 === 0) {
          acc += next;
          console.log("in if - acc:", acc);
        } else {
          acc += 0;
          console.log("in else - acc:", acc);
        }
      }, 0);

      return sum;
    }
    var sum = sumEven(1,2,3,4) // 6
    console.log("sum:", sum);

输出:

next: 1
in else - acc: 0

next: 2
in if - acc: NaN

next: 3
in else - acc: NaN

next: 4
in if - acc: NaN
sum: undefined

Answer 1

您应该在回调函数末尾返回acc

function sumEvenArgs(...args){
    var sum = args.reduce( (acc, next) => {
        console.log("\nnext:", next);
        if (next % 2 === 0) {
            acc += next;
            console.log("in if - acc:", acc);
        } else {
            acc += 0;
            console.log("in else - acc:", acc);
        }
        return acc ; // you need to add this line 
    }, 0);

    return sum;
}
var sum = sumEvenArgs(1,2,3,4) // 6
console.log("sum:", sum);