javascript - 累加器在 make `reduce` 方法中返回未定义

Question

我尝试使函数与 Array.prototype.reduce 方法相同。

function a(collection, iterator, accumulator) {

    for (i of Object.keys(collection)){
      if (i===0 && accumulator === undefined){
          accumulator = collection[i]
          continue
      }

      accumulator = iterator(accumulator, collection[i])
    }

    return accumulator;
  };



a([1,2,3],function(acc,cur){return acc += cur}, 0) // this works fine. returns 6
a([1,2,3],function(acc,cur){return acc += cur}) // expected 6 but returns NaN
a([1,2,3], function(memo){return memo}); // expected 1 but returns undefined

我不知道为什么 accumulator 没有设置为 collection[i] 并在初始值(函数a)的第三个参数未传递。

Answer 1

Object.keys 始终返回一个字符串数组，因此 i 是一个字符串，因此 i === 0 始终为 false。

另请注意，您需要声明i。现在，您的代码正在成为 The Horror of Implicit Globals 的牺牲品。 (我建议使用严格模式，这样它们就应该是错误。)

FWIW，这是解决这两个问题的最低更新(但也请参见下文):

"use strict";
function a(collection, iterator, accumulator) {
    let first = true;                            // ***
    for (const i of Object.keys(collection)){
    //   ^^^^^−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− *** declare `i`
        if (first && accumulator === undefined){
    //      ^^^^^−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− *** first pass?
            first = false;                       // ***
            accumulator = collection[i]
            continue
        }
        first = false;                           // ***

        accumulator = iterator(accumulator, collection[i])
    }

    return accumulator;
};

console.log(a([1,2,3],function(acc,cur){return acc += cur}, 0));
console.log(a([1,2,3],function(acc,cur){return acc += cur}));
console.log(a([1,2,3], function(memo){return memo}));

---

FWIW，我不知道您的意思是否是与 reduce 完全匹配，但如果是这样，您的函数的逻辑与 Array 的逻辑不同。原型(prototype).reduce。请参阅the specification有关详细信息，但 reduce 检查它获得了多少个参数，它不会检查累加器是否未定义。如果您明确为累加器指定值undefined，它将使用该值(而不是使用第一个数组元素的值):

console.log("`undefined` for accumulator:");
["a", "b", "c"].reduce((acc, value) => { console.log(acc, value); return value; }, undefined);

console.log("No accumulator:");
["a", "b", "c"].reduce((acc, value) => { console.log(acc, value); return value; })

.as-console-wrapper {
    max-height: 100% !important;
}

在归约过程中，累加器也可能变得未定义。如果没有给定累加器，则使用第一个看到的值(不一定位于稀疏数组中的索引 0 处!)。 (我对上面累加器的修复实际上修复了这个副产品。)您的代码还使用 Object.keys 而 reduce 仅使用 for 循环。我没有仔细阅读，可能还有其他差异。

"use strict";
function a(collection, iterator, ...optional) {  // ***
    let haveAccumulator = optional.length > 0;   // *** Default acc correctly
    let [accumulator] = optional;                // ***
    for (const i of Object.keys(collection)){
    //   ^^^^^−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− *** declare `i`
        if (!haveAccumulator) {                  // ***
            haveAccumulator = true;              // ***
            accumulator = collection[i]
            continue;
        }

        accumulator = iterator(accumulator, collection[i])
    }

    return accumulator;
};



console.log(a([1,2,3],function(acc,cur){return acc += cur}, 0));
console.log(a([1,2,3],function(acc,cur){return acc += cur}));
console.log(a([1,2,3], function(memo){return memo}));