下面的位运算符在现实世界中有哪些用例?

和 XOR 不 或 左/右转


当前回答

我不认为这是按位计算的,但是ruby的Array通过普通整数按位操作符定义了集合操作。因此[1,2,4]&[1,2,3]# =>[1,2]。对于a ^ b# =>集差值和| b# =>并集也是如此。

其他回答

在当今现代语言的抽象世界里,没有太多。File IO是一个容易想到的方法,尽管它是在已经实现的东西上执行按位操作,而不是实现使用按位操作的东西。尽管如此,作为一个简单的例子,这段代码演示了在c#中删除文件上的只读属性(这样它就可以与指定FileMode.Create的新FileStream一起使用):

//Hidden files posses some extra attibutes that make the FileStream throw an exception
//even with FileMode.Create (if exists -> overwrite) so delete it and don't worry about it!
if(File.Exists(targetName))
{
    FileAttributes attributes = File.GetAttributes(targetName);

    if ((attributes & FileAttributes.ReadOnly) == FileAttributes.ReadOnly)
        File.SetAttributes(targetName, attributes & (~FileAttributes.ReadOnly));

    File.Delete(targetName);
}

As far as custom implementations, here's a recent example: I created a "message center" for sending secure messages from one installation of our distributed application to another. Basically, it's analogous to email, complete with Inbox, Outbox, Sent, etc, but it also has guaranteed delivery with read receipts, so there are additional subfolders beyond "inbox" and "sent." What this amounted to was a requirement for me to define generically what's "in the inbox" or what's "in the sent folder". Of the sent folder, I need to know what's read and what's unread. Of what's unread, I need to know what's received and what's not received. I use this information to build a dynamic where clause which filters a local datasource and displays the appropriate information.

下面是枚举是如何组合在一起的:

    public enum MemoView :int
    {
        InboundMemos = 1,                   //     0000 0001
        InboundMemosForMyOrders = 3,        //     0000 0011
        SentMemosAll = 16,                  //     0001 0000
        SentMemosNotReceived = 48,          //     0011
        SentMemosReceivedNotRead = 80,      //     0101
        SentMemosRead = 144,                //     1001
        Outbox = 272,                       //0001 0001 0000
        OutBoxErrors = 784                  //0011 0001 0000
    }

你明白这是怎么回事了吗?通过与“收件箱”枚举值InboundMemos加上(&),我知道InboundMemosForMyOrders在收件箱中。

下面是该方法的简化版本,它构建并返回为当前选择的文件夹定义视图的过滤器:

    private string GetFilterForView(MemoView view, DefaultableBoolean readOnly)
    {
        string filter = string.Empty;
        if((view & MemoView.InboundMemos) == MemoView.InboundMemos)
        {
            filter = "<inbox filter conditions>";

            if((view & MemoView.InboundMemosForMyOrders) == MemoView.InboundMemosForMyOrders)
            {
                filter += "<my memo filter conditions>";
            }
        }
        else if((view & MemoView.SentMemosAll) == MemoView.SentMemosAll)
        {
            //all sent items have originating system = to local
            filter = "<memos leaving current system>";

            if((view & MemoView.Outbox) == MemoView.Outbox)
            {
                ...
            }
            else
            {
                //sent sub folders
                filter += "<all sent items>";

                if((view & MemoView.SentMemosNotReceived) == MemoView.SentMemosNotReceived)
                {
                    if((view & MemoView.SentMemosReceivedNotRead) == MemoView.SentMemosReceivedNotRead)
                    {
                        filter += "<not received and not read conditions>";
                    }
                    else
                        filter += "<received and not read conditions>";
                }
            }
        }

        return filter;
    }

非常简单,但在抽象级别上是一个整洁的实现,通常不需要按位操作。

还没人提到过收藏。有时您有一个较小的可能值集合,比如只有10或20个可能值,您希望将其中一些值保存在一个集合中。当然,您可以使用常规的Set实现,它很可能使用支持哈希表。但由于可能值的集合是如此之小,这实际上只是浪费时间和空间。相反,您可以将集合存储在单个int或长值中,这正是java EnumSet所做的,如果我没记错的话。

例如,我使用它们从打包的颜色值中获取RGB(A)值。

我一直假设按位操作是相当简单的操作,所以当运行时间至关重要时,通过bitset实现的解决方案可以通过恒定的数量提高运行时间,这取决于算法。

一个数x是2的幂吗?(例如,在计数器递增的算法中很有用,并且一个操作只执行对数次)

(x & (x - 1)) == 0

整数x的最高位是哪位?(例如,这可以用来找出比x大的2的最小次幂)

x |= (x >>  1);
x |= (x >>  2);
x |= (x >>  4);
x |= (x >>  8);
x |= (x >> 16);
return x - (x >>> 1); // ">>>" is unsigned right shift

整数x的最小1位是哪一位?(帮助找出能被2整除的次数。)

x & -x