一元运算与任务融合 [英] Unary Operations fused with assignment

问题描述

可疑的结果是以下代码:

Doubtful result in the following code:

public static void main (String[] args)
{ 
int i = 2;
i = i+=2 + i++;
System.out.println(i); }

期望输出为 8 ，因为'i + = 2'应该 update i，但其行为并非如此.

Was expecting 8 as output, as 'i+=2' should update i, but its not behaving so.

输出:6

Output: 6

我推断速记赋值运算符正在按预期返回 4 ，但没有更新变量i中的相同值.任何解释将不胜感激.

I infer that the short-hand assignment operator is returning 4 as expected but not updating the same in variable i. Any explanation will be appreciated.

推荐答案

i++是后缀增量-它递增i，然后本质上返回i的旧值.等效的前缀运算符++i将返回更新的"值，但这不是这里使用的值.

i++ is a postfix increment - it increments i, then essentially returns the old value of i. The equivalent prefix operator, ++i, would return the "updated" value, but that's not what's being used here.

i+=2的工作原理不同，它基本上等同于i+2，因为它确实返回更新后的值.

i+=2 works differently however, it's essentially equivalent to i+2, since it does return the updated value.

但是，我认为引起混乱的地方是您正在这样看待它:

However, I think where the confusion arises is that you're looking at it like this:

i = (i += 2) + i++;

... 确实给出的预期结果. i+=2给出4，并将i更新为4，然后i++返回4(因为它是后增量，而不是5).但是，当您将运算符优先级放到公式中时，Java实际上像这样括弧"默认情况下:

...which does give your expected result. i+=2 gives 4, and updates i to 4, then i++ returns 4 (instead of 5 since it's a post increment.) However, when you take operator precedence into the equation, Java actually "brackets" it like this by default:

i = i += (2 + i++);

只是为了消除任何混乱，Java以此方式进行评估，因为+=运算符在此示例中优先级最低，因此，首先计算加法表达式(+).

Just to clear up any confusion, Java evaluates it this way because the += operator has least precedence in this example, and therefore the addition expression (+) is calculated first.

此括号内的语句基本上等同于:

This bracketed statement is essentially equivalent to:

i = (i = i + (2 + i++));

依次简化为:

i = i + (2 + i++);

因此，根据上述陈述，并从左至右进行评估，我们首先获取i(2)的值，然后将2+i++的值添加到其中；后者给出4(因为后缀增加).所以我们的最终结果是2 + 4，即6.

So given the above statement, and evaluating from left to right, we first take the value of i (2), and then add to it the value of 2+i++; the latter giving 4 (because of the postfix increment). So our final result is 2+4, which is 6.

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