为什么 [5,6,8,7][1,2] = 8 在 JavaScript 中？
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Why does [5,6,8,7][1,2] = 8 in JavaScript?
提问by Joe
I can't wrap my mind around this quirk.
我无法理解这个怪癖。
[1,2,3,4,5,6][1,2,3]; // 4
[1,2,3,4,5,6][1,2]; // 3
I know [1,2,3] + [1,2] = "1,2,31,2", but I can't find what type or operation is being performed.
我知道[1,2,3] + [1,2] = "1,2,31,2"，但我找不到正在执行的类型或操作。
回答by Lightness Races in Orbit
[1,2,3,4,5,6][1,2,3];
^ ^
| |
array + — array subscript access operation,
where index is `1,2,3`,
which is an expression that evaluates to `3`.
The second [...]cannot be an array, so it's an array subscript operation. And the contents of a subscript operation are not a delimited list of operands, but a single expression.
第二个[...]不能是数组，所以是数组下标操作。并且下标操作的内容不是分隔的操作数列表，而是单个表达式。
Read more about the comma operator here.
在此处阅读有关逗号运算符的更多信息。
回答by Mike Samuel
Because (1,2) == 2. You've stumbled across the comma operator(or simpler explanation here).
因为(1,2) == 2。您偶然发现了逗号运算符（或此处更简单的解释）。
Unless commas appear in a declaration list, parameter list, object or array literal, they act like any other binary operator. x, yevaluates x, then evaluates yand yields that as the result.
除非逗号出现在声明列表、参数列表、对象或数组文字中，否则它们的作用就像任何其他二元运算符。 x, y评估x，然后评估y并产生结果。
回答by Imdad
[1,2,3,4,5,6][1,2,3];
Here the second box i.e. [1,2,3]becomes [3]i.e. the last item so the result will be 4
for example if you keep [1,2,3,4,5,6]in an array
这里的第二个框 ie[1,2,3]成为[3]ie 最后一个项目，因此结果将是 4 例如，如果您保留[1,2,3,4,5,6]在数组中
var arr=[1,2,3,4,5,6];
arr[3]; // as [1,2,3] in the place of index is equal to [3]
similarly
相似地
*var arr2=[1,2,3,4,5,6];
// arr[1,2] or arr[2] will give 3*
But when you place a + operator in between then the second square bracket is not for mentioning index. It is rather another array That's why you get
但是当你在中间放置一个 + 运算符时，第二个方括号不是用于提及索引。它是另一个数组这就是为什么你得到
[1,2,3] + [1,2] = 1,2,31,2
i.e.
IE
var arr_1=[1,2,3];
var arr_2=[1,2];
arr_1 + arr_2; // i.e. 1,2,31,2
Basically in the first case it is used as index of array and in the second case it is itself an array.
基本上在第一种情况下它用作数组的索引，在第二种情况下它本身就是一个数组。
