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简简单单学会C#位运算

时间:2016-06-24 20:15:33      阅读:279      评论:0      收藏:0      [点我收藏+]

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一、理解位运算

要学会位运算,首先要清楚什么是位运算?程序中的所有内容在计算机内存中都是以二进制的形式储存的(即:0或1),位运算就是直接对在内存中的二进制数的每位进行运算操作

二、理解数字进制

上面提到了二进制,除了二进制,我们还有很多的进制,下面列举一些常见的进制

10进制数:0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20 (每位满10进1,同时低位补0)
2进制数:00000,00001,00010,00011,00100,00101,00110,00111,01000,01001,01010,01011,01100,01101,01110,01111,10000,10001,10010,10011,10100 (每位满2进1,同时低位补0)
8进制数:00,01,02,03,04,05,06,07,10,11,12,13,14,15,16,17,20,21,22,23,24 (每位满8进1,同时低位补0)
16进制数:0x00,0x01,0x02,0x03,0x04,0x05,0x06,0x07,0x08,0x09,0x0a,0x0b,0x0c,0x0d,0x0e,0x0f,0x10,0x11,0x12,0x13,0x14 (每位满16进1,10~15由A~F字母表示,同时低位补0)

2进制、8进制、16进制、32进制、64进制等转换成10进制计算方法我得出一个公式:(^表示次方,如:2^2,即2的2次方,8^5即8的5次方)

每位数字转换成10进制时=进制数^(次方)数字索引位(从0开始计算)*数字

计算示例:(注意黑粗体字)

2进制数:10100=2^0*0+2^1*0+2^2*1+2^3*0+2^4*1=0+0+4+0+16=20

8进制数:24=8^0*4+8^1*2=4+16=20

16进制数:0x14(注意0x是用来表示16进制数的意思,不是数字本身的内容)=16^0*4+16^1*1=4+16=20

至于各进制之间的转换,比如:2进制转换成16进制,如果想自己手算,一般都是先转成10进制,然后将数字进行与进制数相除,直到得出余数小于或等于进制数(或0),当然作为程序员的我们,应该使用现有的方法,如下:

Convert.ToString(数字,进制数)

如:Convert.ToString(10,2)=01010,Convert.ToString(10,8)=12 ,Convert.ToString(13,16)=0x0d

综合示例如下:

            int i10 = 68;
            int i16 = 0x2A;

            Console.WriteLine("示例一:");

            Console.Write("10进制【68】转成2、8、16进制结果:{0}、{1}、{2}\n",
                        Convert.ToString(i10, 2), Convert.ToString(i10, 8), Convert.ToString(i10, 16));

            Console.Write("16进制【0x2A】转成2、8、10进制结果:{0}、{1}、{2}\n",
            Convert.ToString(i16, 2), Convert.ToString(i16, 8), Convert.ToString(i16, 10));

输出结果:

10进制【68】转成2、8、16进制结果:1000100、104、44
16进制【0x2A】转成2、8、10进制结果:101010、52、42

三、初识位运算(位与与位或运算)

本文一开始就说明了,位运算就是二进制每位数字的运算操作,下面通过代码示例来初识位运算

            Console.WriteLine("示例二:");

            int b0 = 0, b1 = 1, b2 = 2, b3 = 4, b4 = 8, b5 = 16;

            FormatWrite("b0", "b1", b0, b1, "&");
            FormatWrite("b0", "b1", b0, b1, "|");

            Console.WriteLine();

            FormatWrite("b2", "b3", b2, b3, "&");
            FormatWrite("b2", "b3", b2, b3, "|");

            Console.WriteLine();

            FormatWrite("b4", "b5", b4, b5, "&");
            FormatWrite("b4", "b5", b4, b5, "|");




        static void FormatWrite(string n1, string n2, int d1, int d2, string opt)
        {
            string writeMsg = string.Format("{0} {1} {2}", n1, opt, n2);
            writeMsg += string.Format(" = {0} {1} {2}", d1, opt, d2);
            string d1str = Convert.ToString(d1, 2), d2str = Convert.ToString(d2, 2);
            int maxLen = Math.Max(d1str.Length, d2str.Length);
            writeMsg += string.Format(" = {0} {1} {2}", d1str.PadLeft(maxLen, ‘0‘), opt, d2str.PadLeft(maxLen, ‘0‘));
            switch (opt)
            {
                case "&":
                    {
                        writeMsg += string.Format(" = 10进制:{0} 或 2进制:{1}", Convert.ToString(d1 & d2, 10), Convert.ToString(d1 & d2, 2).PadLeft(maxLen, ‘0‘));
                        break;
                    }
                case "|":
                    {
                        writeMsg += string.Format(" = 10进制:{0} 或 2进制:{1}", Convert.ToString(d1 | d2, 10), Convert.ToString(d1 | d2, 2).PadLeft(maxLen, ‘0‘));
                        break;
                    }
            }

            Console.WriteLine(writeMsg);
        }

输出结果:

b0 & b1 = 0 & 1 = 0 & 1 = 10进制:0 或 2进制:0
b0 | b1 = 0 | 1 = 0 | 1 = 10进制:1 或 2进制:1

b2 & b3 = 2 & 4 = 010 & 100 = 10进制:0 或 2进制:000
b2 | b3 = 2 | 4 = 010 | 100 = 10进制:6 或 2进制:110

b4 & b5 = 8 & 16 = 01000 & 10000 = 10进制:0 或 2进制:00000
b4 | b5 = 8 | 16 = 01000 | 10000 = 10进制:24 或 2进制:11000

位与运算:

参加运算的两个数字,按二进制进行与运算,如果两个相应的二进位数为1,则该位的结果为 1, 否则为 0 ,即:
0 & 0 = 0;0 & 1 = 0;1 & 0 = 0;1& 1 = 1 

也就是只有1 & 1才会得1,否则都为0;

位或运算:

参加运算的两个数字,按二进制进行或运算,如果两个相应的二进位中只要有一个为 1,则该位的结果就为 1,否则为 0 ,即:

0|0=0; 0|1=1; 1|0=1; 1|1=1;

也就是只有0 & 0才会得0,否则都为1;

 四、寻找规律

我们先看一下10进制及2进制0~20的数字

10进制数:0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20 
2进制数:00000,00001,00010,00011,00100,00101,00110,00111,01000,01001,01010,01011,01100,01101,01110,01111,10000,10001,10010,10011,10100 

从2进制数0~20中,我们发现只要是2的偶数次方时,则位数发生变化,且多位中只有一个为1,其余位均为0,找出的数字如下:

00000、00001、00010、,00100、01000、10000

对应10进制数:0、1、2、4、8、16

如果对这些数全部进行位或运算,则最终的结果是:11111,即5位会部是1

从这里发现了什么呢?我是看出来了,不知道各位看官是否看出规律,其实很简单,如果我们把每一位都当成一个控制开关或者说是存在不存在,那么0表示关或者不存在,1表示开或者存在,那么我们可以针对这个规律实现复杂的组合控制。

其实微软早就应用了这个规律特性,比如:

typeof(Program).GetProperties(BindingFlags.Instance | BindingFlags.Public | BindingFlags.Static);

BindingFlags定义如下:

技术分享

如上,每一个枚举项都是2的次方,每一项都是上一项的2倍,我们也可以利用这个规律实现类似的处理。

五、实现组合控制

            Console.WriteLine("示例四:");

            ButtonStyle userbtnStyle = ButtonStyle.OK | ButtonStyle.Cancel | ButtonStyle.Alert | ButtonStyle.Info;//用户需要显示的ButtonStyle,通过或运算组合在一起,得出2进制值:1111
            string buttonStyleStr = null;


            //进行位逻辑判断,能够准确识别userbtnStyle的组合的内容
            if ((userbtnStyle & ButtonStyle.AlertInfo) == ButtonStyle.AlertInfo) 
            {
                buttonStyleStr += "+" + Enum.GetName(typeof(ButtonStyle), ButtonStyle.AlertInfo);
            }
            else if ((userbtnStyle & ButtonStyle.Alert) == ButtonStyle.Alert)
            {
                buttonStyleStr += "+" + Enum.GetName(typeof(ButtonStyle), ButtonStyle.Alert);
            }
            else if ((userbtnStyle & ButtonStyle.Info) == ButtonStyle.Info)
            {
                buttonStyleStr += "+" + Enum.GetName(typeof(ButtonStyle), ButtonStyle.Info);
            }

            if ((userbtnStyle & ButtonStyle.OKCancel) == ButtonStyle.OKCancel)
            {
                buttonStyleStr += "+" + Enum.GetName(typeof(ButtonStyle), ButtonStyle.OKCancel);
            }
            else if ((userbtnStyle & ButtonStyle.OK) == ButtonStyle.OK)
            {
                buttonStyleStr += "+" + Enum.GetName(typeof(ButtonStyle), ButtonStyle.OK);
            }
            else if ((userbtnStyle & ButtonStyle.Cancel) == ButtonStyle.Cancel)
            {
                buttonStyleStr += "+" + Enum.GetName(typeof(ButtonStyle), ButtonStyle.Cancel);
            }

            Console.WriteLine("需要显示的按钮有:" + buttonStyleStr.Substring(1));




        enum ButtonStyle
        {
            None = 0x00,
            OK = 0x01,
            Cancel = 0x02,
            Alert = 0x04,
            Info = 0x08,
            OKCancel = 0x01 | 0x02,
            AlertInfo = 0x04 | 0x08
        }

输出结果:

需要显示的按钮有:AlertInfo+OKCancel

如果改变userbtnStyle的组合,得到的结果也会不同

另外一个示例:

            Console.WriteLine("示例五:");

            AllowType userPermission = AllowType.Add | AllowType.Update | AllowType.Upload | AllowType.Download | AllowType.Select;
            string userPermissionStr = null;

            if ((userPermission & AllowType.Edit) == AllowType.Edit)
            {
                userPermissionStr += "+" + Enum.GetName(typeof(AllowType), AllowType.Edit);
            }
            else
            {
                if ((userPermission & AllowType.Add) == AllowType.Add)
                {
                    userPermissionStr += "+" + Enum.GetName(typeof(AllowType), AllowType.Add);
                }

                if ((userPermission & AllowType.Update) == AllowType.Update)
                {
                    userPermissionStr += "+" + Enum.GetName(typeof(AllowType), AllowType.Update);
                }

                if ((userPermission & AllowType.Delete) == AllowType.Delete)
                {
                    userPermissionStr += "+" + Enum.GetName(typeof(AllowType), AllowType.Delete);
                }

                if ((userPermission & AllowType.Upload) == AllowType.Upload)
                {
                    userPermissionStr += "+" + Enum.GetName(typeof(AllowType), AllowType.Upload);
                }

            }

            if ((userPermission & AllowType.Read) == AllowType.Read)
            {
                userPermissionStr += "+" + Enum.GetName(typeof(AllowType), AllowType.Read);
            }
            else
            {

                if ((userPermission & AllowType.Select) == AllowType.Select)
                {
                    userPermissionStr += "+" + Enum.GetName(typeof(AllowType), AllowType.Select);
                }


                if ((userPermission & AllowType.Download) == AllowType.Download)
                {
                    userPermissionStr += "+" + Enum.GetName(typeof(AllowType), AllowType.Download);
                }
            }

            Console.WriteLine("用户具备的权限有:" + userPermissionStr.Substring(1));


        enum AllowType
        {
            None = 0,
            Add = 1,
            Update = 2,
            Delete = 4,
            Select = 8,
            Upload = 16,
            Download = 32,
            Edit = Add | Update | Delete | Upload,
            Read = Select | Download
        }

输出结果:

用户具备的权限有:Add+Update+Upload+Read

如果改变userPermission的组合,得到的结果也会不同

上述两个例子,就是允分利用位或位与运算,大家可以理解位或是将两者拼在一起,位与是找出组合中是否有包含的部份

六、了解其它位运算

            Console.WriteLine("示例六:");

            int x1 = 108;
            Console.Write("~位非运算:{0} -->> {1} ; {2} -->> {3}\n",
                        Convert.ToString(x1, 10), Convert.ToString(~x1, 10),
                        Convert.ToString(x1, 2), Convert.ToString(~x1, 2));

            Console.Write("<<位左移(移5位)运算:{0} -->> {1} ; {2} -->> {3}\n",//说白了讲:左移N位就是将二进制数后面补N位0
            Convert.ToString(x1, 10), Convert.ToString(x1 << 5, 10),
            Convert.ToString(x1, 2), Convert.ToString(x1 << 5, 2));

            Console.Write(">>位右移(移5位)运算:{0} -->> {1} ; {2} -->> {3}\n",//说白了讲:右移N位就是将二进制数后面删除N位数(不论0或1)
            Convert.ToString(x1, 10), Convert.ToString(x1 >> 5, 10),
            Convert.ToString(x1, 2), Convert.ToString(x1 >> 5, 2));

            Console.Write("^位异或(异或5)运算:{0} ^ {1} -->> {2} ; {3} ^ {4}-->> {5}\n",//说白了讲:右移N位就是将二进制数后面删除N位数(不论0或1)
            Convert.ToString(x1, 10),Convert.ToString(5, 10), Convert.ToString(x1 ^ 5, 10),
            Convert.ToString(x1, 2), Convert.ToString(5, 2), Convert.ToString(x1 ^ 5, 2));

输出结果:

~位非运算:108 -->> -109 ; 1101100 -->> 11111111111111111111111110010011
<<位左移(移5位)运算:108 -->> 3456 ; 1101100 -->> 110110000000
>>位右移(移5位)运算:108 -->> 3 ; 1101100 -->> 11
^位异或(异或5)运算:108 ^ 5 -->> 105 ; 1101100 ^ 101-->> 1101001

 

~非运算:是一个单项运算符,用来对一个二进制按位取反,即将 0 变 1,1变 0。

<<左移:用来对一个数每个二进位全部左移若干位,说白了讲:左移N位就是将二进制数后面补N位0

>>右移:用来对一个数每个二进位全部右移若干位,移到右端的低位被舍弃,对无符号数,高位补 0,说白了讲:右移N位就是将二进制数后面删除N位数(不论0或1)

^异或运算: 也称 XOR 运算符。它的规则是若参加运算的两个二进位同号,则结果为0,异号则为1。即 0^0=0; 0^1=1; 1^0=1;1^1=0;说白了讲:若两个都为0,则为0,否则相同的则为0,不相同的则为1

 

为了便于大家进行各种测试,贴出DEMO代码,供大家学习:

using System;
using System.Collections.Generic;
using System.Linq;
using System.Reflection;
using System.Text;

namespace TestConsoleApp
{
    class Program
    {
        static void Main(string[] args)
        {
            Console.SetBufferSize(800, 600);
            Console.WriteLine("数据进制了解:");
            int[] nums = new[] { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20 };
            Console.WriteLine("10进制数:" + string.Join(",", nums));
            Console.WriteLine("2进制数:" + GetFormatNumbersToStr(Convert.ToString, 2, true, nums));
            Console.WriteLine("8进制数:" + GetFormatNumbersToStr(Convert.ToString, 8, true, nums));
            Console.WriteLine("16进制数:" + GetFormatNumbersToStr(Convert.ToString, 16, true, nums));
            Console.WriteLine();

            int i10 = 68;//10进制数:68
            int i16 = 0x2A;//16进制=16^0*10+16^1*2=10+32=42   -->相当于10进制数:42

            Console.WriteLine("示例一:");

            Console.Write("10进制【68】转成2、8、16进制结果:{0}、{1}、{2}\n",
                        Convert.ToString(i10, 2), Convert.ToString(i10, 8), Convert.ToString(i10, 16));

            Console.Write("16进制【0x2A】转成2、8、10进制结果:{0}、{1}、{2}\n",
            Convert.ToString(i16, 2), Convert.ToString(i16, 8), Convert.ToString(i16, 10));

            Console.WriteLine();
            Console.WriteLine("示例二:");

            int b0 = 0, b1 = 1, b2 = 2, b3 = 4, b4 = 8, b5 = 16;

            FormatWrite("b0", "b1", b0, b1, "&");
            FormatWrite("b0", "b1", b0, b1, "|");

            Console.WriteLine();

            FormatWrite("b2", "b3", b2, b3, "&");
            FormatWrite("b2", "b3", b2, b3, "|");

            Console.WriteLine();

            FormatWrite("b4", "b5", b4, b5, "&");
            FormatWrite("b4", "b5", b4, b5, "|");


            FormatWrite("0~", "~16", 0 | 1 | 2 | 4, 8|16, "|");

            Console.WriteLine();

            Console.WriteLine("示例三:");

            List<int> dds = new List<int>();
            int d = 0;
            while (d <= 500)
            {
                if (d < 2)
                {
                    dds.Add(d);
                    ++d;
                }
                else
                {
                    dds.Add(d);
                    d = d * 2;
                }
            }

            Console.WriteLine("10进制数:" + GetFormatNumbersToStr(Convert.ToString, 10, true, dds.ToArray()));
            Console.WriteLine("16进制数:" + GetFormatNumbersToStr(Convert.ToString, 16, true, dds.ToArray()));
            Console.WriteLine("2进制数:" + GetFormatNumbersToStr(Convert.ToString, 2, true, dds.ToArray()));


            Console.WriteLine();

            Console.WriteLine("示例四:");

            ButtonStyle userbtnStyle = ButtonStyle.OK | ButtonStyle.Cancel | ButtonStyle.Alert | ButtonStyle.Info;//用户需要显示的ButtonStyle,通过或运算组合在一起,得出2进制值:1111
            string buttonStyleStr = null;


            //进行位逻辑判断,能够准确识别userbtnStyle的组合的内容
            if ((userbtnStyle & ButtonStyle.AlertInfo) == ButtonStyle.AlertInfo) 
            {
                buttonStyleStr += "+" + Enum.GetName(typeof(ButtonStyle), ButtonStyle.AlertInfo);
            }
            else if ((userbtnStyle & ButtonStyle.Alert) == ButtonStyle.Alert)
            {
                buttonStyleStr += "+" + Enum.GetName(typeof(ButtonStyle), ButtonStyle.Alert);
            }
            else if ((userbtnStyle & ButtonStyle.Info) == ButtonStyle.Info)
            {
                buttonStyleStr += "+" + Enum.GetName(typeof(ButtonStyle), ButtonStyle.Info);
            }

            if ((userbtnStyle & ButtonStyle.OKCancel) == ButtonStyle.OKCancel)
            {
                buttonStyleStr += "+" + Enum.GetName(typeof(ButtonStyle), ButtonStyle.OKCancel);
            }
            else if ((userbtnStyle & ButtonStyle.OK) == ButtonStyle.OK)
            {
                buttonStyleStr += "+" + Enum.GetName(typeof(ButtonStyle), ButtonStyle.OK);
            }
            else if ((userbtnStyle & ButtonStyle.Cancel) == ButtonStyle.Cancel)
            {
                buttonStyleStr += "+" + Enum.GetName(typeof(ButtonStyle), ButtonStyle.Cancel);
            }

            Console.WriteLine("需要显示的按钮有:" + buttonStyleStr.Substring(1));



            Console.WriteLine();

            Console.WriteLine("示例五:");

            AllowType userPermission = AllowType.Add | AllowType.Update | AllowType.Upload | AllowType.Download | AllowType.Select;
            string userPermissionStr = null;

            if ((userPermission & AllowType.Edit) == AllowType.Edit)
            {
                userPermissionStr += "+" + Enum.GetName(typeof(AllowType), AllowType.Edit);
            }
            else
            {
                if ((userPermission & AllowType.Add) == AllowType.Add)
                {
                    userPermissionStr += "+" + Enum.GetName(typeof(AllowType), AllowType.Add);
                }

                if ((userPermission & AllowType.Update) == AllowType.Update)
                {
                    userPermissionStr += "+" + Enum.GetName(typeof(AllowType), AllowType.Update);
                }

                if ((userPermission & AllowType.Delete) == AllowType.Delete)
                {
                    userPermissionStr += "+" + Enum.GetName(typeof(AllowType), AllowType.Delete);
                }

                if ((userPermission & AllowType.Upload) == AllowType.Upload)
                {
                    userPermissionStr += "+" + Enum.GetName(typeof(AllowType), AllowType.Upload);
                }

            }

            if ((userPermission & AllowType.Read) == AllowType.Read)
            {
                userPermissionStr += "+" + Enum.GetName(typeof(AllowType), AllowType.Read);
            }
            else
            {

                if ((userPermission & AllowType.Select) == AllowType.Select)
                {
                    userPermissionStr += "+" + Enum.GetName(typeof(AllowType), AllowType.Select);
                }


                if ((userPermission & AllowType.Download) == AllowType.Download)
                {
                    userPermissionStr += "+" + Enum.GetName(typeof(AllowType), AllowType.Download);
                }
            }

            Console.WriteLine("用户具备的权限有:" + userPermissionStr.Substring(1));


            Console.WriteLine();

            Console.WriteLine("示例六:");

            int x1 = 108;
            Console.Write("~位非运算:{0} -->> {1} ; {2} -->> {3}\n",
                        Convert.ToString(x1, 10), Convert.ToString(~x1, 10),
                        Convert.ToString(x1, 2), Convert.ToString(~x1, 2));

            Console.Write("<<位左移(移5位)运算:{0} -->> {1} ; {2} -->> {3}\n",//说白了讲:左移N位就是将二进制数后面补N位0
            Convert.ToString(x1, 10), Convert.ToString(x1 << 5, 10),
            Convert.ToString(x1, 2), Convert.ToString(x1 << 5, 2));

            Console.Write(">>位右移(移5位)运算:{0} -->> {1} ; {2} -->> {3}\n",//说白了讲:右移N位就是将二进制数后面删除N位数(不论0或1)
            Convert.ToString(x1, 10), Convert.ToString(x1 >> 5, 10),
            Convert.ToString(x1, 2), Convert.ToString(x1 >> 5, 2));

            Console.Write("^位异或(异或5)运算:{0} ^ {1} -->> {2} ; {3} ^ {4}-->> {5}\n",//说白了讲:右移N位就是将二进制数后面删除N位数(不论0或1)
            Convert.ToString(x1, 10),Convert.ToString(5, 10), Convert.ToString(x1 ^ 5, 10),
            Convert.ToString(x1, 2), Convert.ToString(5, 2), Convert.ToString(x1 ^ 5, 2));

            Console.ReadKey();
        }

        static void FormatWrite(string n1, string n2, int d1, int d2, string opt)
        {
            string writeMsg = string.Format("{0} {1} {2}", n1, opt, n2);
            writeMsg += string.Format(" = {0} {1} {2}", d1, opt, d2);
            string d1str = Convert.ToString(d1, 2), d2str = Convert.ToString(d2, 2);
            int maxLen = Math.Max(d1str.Length, d2str.Length);
            writeMsg += string.Format(" = {0} {1} {2}", d1str.PadLeft(maxLen, ‘0‘), opt, d2str.PadLeft(maxLen, ‘0‘));
            switch (opt)
            {
                case "&":
                    {
                        writeMsg += string.Format(" = 10进制:{0} 或 2进制:{1}", Convert.ToString(d1 & d2, 10), Convert.ToString(d1 & d2, 2).PadLeft(maxLen, ‘0‘));
                        break;
                    }
                case "|":
                    {
                        writeMsg += string.Format(" = 10进制:{0} 或 2进制:{1}", Convert.ToString(d1 | d2, 10), Convert.ToString(d1 | d2, 2).PadLeft(maxLen, ‘0‘));
                        break;
                    }
            }

            Console.WriteLine(writeMsg);
        }

        static string GetFormatNumbersToStr(Func<int, int, string> ConvertToStringFunc, int toBase, bool showH, params int[] nums)
        {
            List<string> strs = nums.Select(n => ConvertToStringFunc(n, toBase)).ToList();
            int maxLen = strs.Max(s => s.Length);
            string strLine = null;
            foreach (string str in strs)
            {
                string str1 = str.PadLeft(maxLen, ‘0‘);
                if (toBase == 16) str1 = "0x" + str1;

                if (showH)
                {
                    strLine += "," + str1;
                }
                else
                {
                    strLine += ",\n" + str1;
                }
            }
            return strLine.Substring(1);
        }

        enum ButtonStyle
        {
            None = 0x00,
            OK = 0x01,
            Cancel = 0x02,
            Alert = 0x04,
            Info = 0x08,
            OKCancel = 0x01 | 0x02,
            AlertInfo = 0x04 | 0x08
        }


        enum AllowType
        {
            None = 0,
            Add = 1,
            Update = 2,
            Delete = 4,
            Select = 8,
            Upload = 16,
            Download = 32,
            Edit = Add | Update | Delete | Upload,
            Read = Select | Download
        }


    }
}

  

简简单单学会C#位运算

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原文地址:http://www.cnblogs.com/zuowj/p/5614237.html

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