累加器 returns 在 making `reduce` 方法中未定义

Question

我试图让一个功能与 Array.prototype.reduce 方法一样工作。

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在初始值（函数a的第三个参数）未传递时不设置为collection[i]并保持undefined。

Answer 1

Object.keys总是returns一个字符串数组，所以i是一个字符串，所以i === 0总是假的.

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，这样他们就应该出现错误。）

FWIW，这是解决这两个问题的最小更新（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，我不知道你的意思是不是 reduce 的完全匹配，但如果是这样，你的函数逻辑与 Array.prototype.reduce 的逻辑不同。有关详细信息，请参阅 the specification，但 reduce 检查它获得了多少个参数，它不会检查累加器是否为 undefined。如果您明确地为累加器指定值 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;
}

累加器在减法的时候也可以变成undefined。如果没有给出累加器，则使用第一个看到的值（不一定是稀疏数组中的索引 0！）。（我对上述累加器的修复实际上将其修复为 by-product。）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}));

累加器 returns 在 making `reduce` 方法中未定义

accumulator returns undefined within making `reduce` method

javascript

reduce