了解增量 [英] Understanding Incrementing
问题描述
例如：
var a = 123;
var b = a++;
现在 a 包含 124 和 b 包含 123
我知道b取的是a的值，然后a正在递增。但是，我不明白为什么会这样。为什么JavaScript的创建者会想要这个的主要原因。除了混淆新手之外，有什么好处？
I understand that b is taking the value of a and then a is being incremented. However, I don't understand why this is so. The principal reason for why the creators of JavaScript would want this. What is the advantage to this other than confusing newbies?
推荐答案
这就是为什么它被称为后递增运算符。基本上，一切都是表达式，它会产生一个值。 a + 1 是一个表达式，其结果为124.如果您使用<$ c $将此值分配给 b c> b = a + 1 ， b 的值为124.如果您没有将结果分配给任何内容， a + 1 仍然会产生值124，它会立即丢弃，因为你没有在任何地方捕捉它。
That's why it's called the "post-incrementing operator". Essentially, everything is an expression which results in a value. a + 1 is an expression which results in the value 124. If you assign this to b with b = a + 1, b has the value of 124. If you do not assign the result to anything, a + 1 will still result in the value 124, it will just be thrown away immediately since you're not "catching" it anywhere.
BTW，偶数 b = a + 1 是一个返回124的表达式。赋值表达式的结果值是赋值。这就是为什么 c = b = a + 1 按预期工作的原因。
BTW, even b = a + 1 is an expression which returns 124. The resulting value of an assignment expression is the assigned value. That's why c = b = a + 1 works as you'd expect.
无论如何，关于这一点的特别之处带有 ++ 和 - 的表达式除了返回值之外， ++ 运算符直接修改变量。那么当你执行 b = a ++ 时会发生什么，表达式 a ++ 返回值123并递增 A 。 帖子 递增器首先返回值 然后 增量，而 pre incrementor ++ a 首先 增量，然后返回该值。如果您在没有作业的情况下单独编写 a ++ ，您将不会注意到差异。通常使用 a ++ ，作为的短手a = a + 1 。
Anyway, the special thing about an expression with ++ and -- is that in addition to returning a value, the ++ operator modifies the variable directly. So what happens when you do b = a++ is, the expression a++ returns the value 123 and increments a. The post incrementor first returns the value, then increments, while the pre incrementor ++a first increments, then returns the value. If you just wrote a++ by itself without assignment, you won't notice the difference. That's how a++ is usually used, as short-hand for a = a + 1.
这是非常标准的。
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