表达式“ p [i& 1] + = v，p”是什么意思？ [英] What does the expression `p[i&1]+=v,p` mean?

问题描述

我正在使用CodeFights练习javascript，完成练习后，我看到了此功能的结果：

I was practicing my javascript with CodeFights and after I finished an exercise I saw this function as a result:

// Subject : 
// Several people are standing in a row and need to be divided into two teams. 
// The first person goes into team 1, the second goes into team 2, 
// the third goes into team 1 again, the fourth into team 2, and so on.
// You are given an array of positive integers - the weights of the people. 
// Return an array of two integers, where the first element is the total weight of 
// team 1, and the second element is the total weight of team 2 
// after the division is complete.

// Example :
// For a = [50, 60, 60, 45, 70], the output should be
// alternatingSums(a) = [180, 105].

// answer
alternatingSums = a => a.reduce((p,v,i) => (p[i&1]+=v,p), [0,0])

我不明白 p [i& 1] + = v，p 是什么意思。

推荐答案

& 符号是按位二进制运算符。

The & symbol is a bitwise binary operator.

要理解会发生什么，您必须将每个项目转换为二进制。

To understand what would happen, you have to convert each item to binary.

   | i (decimal) | i (binary) | i & 1 |
   |-------------|------------|-------|
   |           0 |          0 |     0 |
   |           1 |          1 |     1 |
   |           2 |         10 |     0 |
   |           3 |         11 |     1 |
   |           4 |        100 |     0 |
   |           5 |        101 |     1 |

有效地，每个偶数将转换为0，每个奇数将转换为1。

Effectively, every even number will be transformed to 0, and every odd number will be transformed to 1.

如果我想实现这一目标，我个人将使用模运算符（％）

If I was trying to achieve that outcome, I personally would have used the modulus operator (%)

 p[i%2] += v;

但这就是我。

< hr />
另一部分是用逗号分隔两个语句：

---

The other part is that there are two statements separated by a comma:

 (p[i&1]+=v,p)

这是说执行此操作，然后返回 p 。它的简写为：

That's saying "Perform this action, then return p. It's shorthand for:

alternatingSums = a => a.reduce((p,v,i) => { 
                                              p[i&1]+=v;
                                              return p;
                                           }, 
                                [0,0])

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