FLASH与密码学-常用加密算法的AS3.0实现
如果对技术有兴趣的，以下这编值得看看起来
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DES加密算法
DES是一种相对”古老”的算法，在上个世纪70年代由IBM开发完成。在1976年被美国政府采用，并于1977年作为非机要部门的数据加密标准而公布。DES是一种单密钥对称加解密算法。
DES算法流程
DES是一个分组加密算法，利用一个64位的密钥对64位的数据进行加密，输出64位的密文,64位的密钥当中，每一个字节的第8位作为奇偶校验位而忽略，所以实际上的密钥是56位，DES算法的步骤如下：
初始明文X经过初始IP置换表得到IP(X), 把IP(X)分为两部分(L0,R0)，经过16轮置换,XOR和替换操作后,通过逆IP置换表生成密文,公式如下：
Ln = R(n-1)
Rn = L(n-1) + f (R(n-1),Kn)
+号表示XOR异或操作。Kn为密钥分组。整个加密流程见下图
生成子密钥
K1-K16是由64位密钥经过16轮的循环移位生成的16组子密钥。
设密钥K（64位）= 01100011 01101111 01101101 01110000 01110101 01110100 01100101 01110010（大家经过8421码算一下，转换成16进制就是63 6F 6D 70 75 74 65 72,其实就是字符串computer）。注意红色标识的为奇偶校验位，实际密钥位56位。
对K使用PC-1表进行置换：
得出K+ (56位)= 00000000 11111111 11111111 10111000 00110111 01100000 01101000
把K+均分为两部分，那么
C0??=??00000000 11111111 11111111 1011 ，
D0 =? ?1000 00110111 01100000 01101000
然后开始根据移位常数表进行16轮移位，第1，2，9，16轮循环左移1位，其他左移2位。
那么
C1 = C0 循环左移1位 = 0000 0001 1111 1111 1111 1111 0110
D1 = C1 循环左移1位 = 0000 0110 1110 1100 0000 1101 0001
C2 = C1 循环左移1位 = 0000 0011 1111 1111 1111 1110 1100
D2 = C2 循环左移1位 = 0000 1101 1101 1000 0001 1010 0010
…….
C16 = C15 循环左移1位 = 0000 0000 1111 1111 1111 1111 1011
D16 = D15 循环左移1位 = 1000 0011 0111 0110 0000 0110 1000? ? ? ?? ? ? ?
C(n) + D(n) = K(n)， 对Kn使用PC-2表进行置换，得出16组子密钥(48位1组)
K1??= 11110000 10111110 11101110 11010000 00000111 10011000
……
K16 = 11110001 10111110 00101110 00000001 10000010 01011110
加密过程
初始置换
设明文M(64位) = 01110100 01100101 01110011 01110100 00100000 01100100 01100101 01110011(就是字符串test des)
对明文M使用IP表进行置换
IP??=??11101111 10001101 01101011 11000110 00000000 11111111 00000000 10000100
进而
L0 =??11101111 10001101 01101011 11000110
R0 =??00000000 11111111 00000000 10000100
16轮变换
从L0,R0开始 循环16次，使用上文的公式
Ln = R(n-1)
Rn = L(n-1) + f (R(n-1),Kn)
得出L1R1-L16R16,??函数f从一个32位的数据块和一个48位的子密钥Kn得到一个新的32位数据块(算法稍后详细介绍) 最终得出
L16 = 11000000 10110011 00110001 10001011
R16 = 10101110 10111011 11100010 10110010
对R16L16(64位)使用IP-1表进行逆置换，得出密文:
00111010 01110111 01000000 01010010 00111001 01111101 10000100 11110111
最终我们从明文M??= 74 65 73 74 20 64 65 73
得出密文? ?? ? C??= 3A 77 40 52 39 7D 84 F7
DES的加密与解密使用相同的算法，把密钥的使用顺序颠倒即可。
下面开始正式的编码
使用FD创建一个空项目，新建一个名为DES的类,方便起见我们把DES设成单件类,
第一步把我们上面所讲到的一些表编码进类里面
package??
{
????/**
???? * ...
???? * @author bardpub
???? */
????public class DES
????{
????????
????????private static var _instance:DES;
????????
????????// 置换选择表一 (PC-1)
????????public static var bSelSwapTable_1:Array = [
????????????57, 49, 41, 33, 25, 17,??9,??1, 58, 50, 42, 34, 26, 18,
????????????10,??2, 59, 51, 43, 35, 27, 19, 11,??3, 60, 52, 44, 36,
????????????63, 55, 47, 39, 31, 23, 15,??7, 62, 54, 46, 38, 30, 22,
????????????14,??6, 61, 53, 45, 37, 29, 21, 13,??5, 28, 20, 12,??4
????????] ;
????????// 置换选择表二 (PC-2)
????????public static var bSelSwapTable_2:Array = [
????????????14, 17, 11, 24,??1,??5,??3, 28, 15,??6, 21, 10,
????????????23, 19, 12,??4, 26,??8, 16,??7, 27, 20, 13,??2,
????????????41, 52, 31, 37, 47, 55, 30, 40, 51, 45, 33, 48,
????????????44, 49, 39, 56, 34, 53, 46, 42, 50, 36, 29, 32
????????] ;
????????
????????// 初始置换表 (IP)
????????public static var??bInitSwapTable:Array = [
????????????58, 50, 42, 34, 26, 18, 10, 2, 60, 52, 44, 36, 28, 20, 12, 4,
????????????62, 54, 46, 38, 30, 22, 14, 6, 64, 56, 48, 40, 32, 24, 16, 8,
????????????57, 49, 41, 33, 25, 17, 9,??1, 59, 51, 43, 35, 27, 19, 11, 3,
????????????61, 53, 45, 37, 29, 21, 13, 5, 63, 55, 47, 39, 31, 23, 15, 7
????????] ;
????????// 初始逆置换表 (IP-1)
????????public static var bInitReSwapTable:Array = [
????????????40, 8, 48, 16, 56, 24, 64, 32, 39, 7, 47, 15, 55, 23, 63, 31,
????????????38, 6, 46, 14, 54, 22, 62, 30, 37, 5, 45, 13, 53, 21, 61, 29,
????????????36, 4, 44, 12, 52, 20, 60, 28, 35, 3, 43, 11, 51, 19, 59, 27,
????????????34, 2, 42, 10, 50, 18, 58, 26, 33, 1, 41,??9, 49, 17, 57, 25
????????] ;
????????
????????// 位扩展表32-->48 (E)
????????public static var bBitExternTable:Array = [
????????????32,??1,??2,??3,??4,??5,??4,??5,??6,??7,??8,??9,
????????????8,??9, 10, 11, 12, 13, 12, 13, 14, 15, 16, 17,
????????????16, 17, 18, 19, 20, 21, 20, 21, 22, 23, 24, 25,
????????????24, 25, 26, 27, 28, 29, 28, 29, 30, 31, 32,??1
????????] ;
????????// 32位置换表--用于F的尾置换 (P)
????????public static var bTailSwapTable:Array = [
????????????16,??7, 20, 21, 29, 12, 28, 17,
????????????1, 15, 23, 26,??5, 18, 31, 10,
????????????2,??8, 24, 14, 32, 27,??3,??9,
????????????19, 13, 30,??6, 22, 11,??4, 25
????????] ;
????????
????????//8个S盒
????????public static var SB:Array = [
????????[
????????????14, 4, 13, 1, 1, 15, 11, 8, 3, 10, 6, 12, 5, 9, 0, 7,
????????????0, 15, 7, 4, 14, 2, 13, 1, 10, 6, 12, 11, 9, 5, 3, 8,
????????????4, 1, 14, 8, 13, 6, 2, 11, 15, 12, 9, 7, 3, 10, 5, 0,
????????????15, 12, 8, 2, 4, 9, 1, 7, 5, 11, 3, 14, 10, 0, 6, 13
????????],
????????
????????[
????????????15, 1, 8, 14, 6, 11, 3, 4, 9, 7, 2, 13, 12, 0, 5, 10,
????????????3, 13, 4, 7, 15, 2, 8, 14, 12, 0, 1, 10, 6, 9, 11, 5,
????????????0, 14, 7, 11, 10, 4, 13, 1, 5, 8, 12, 6, 9, 3, 2, 15,
????????????13, 8, 10, 1, 3, 15, 4, 2, 11, 6, 7, 12, 0, 5, 14, 9
????????],
????
????????[
????????????10, 0, 9, 14, 6, 3, 15, 5, 1, 13, 12, 7, 11, 4, 2, 8,
????????????13, 7, 0, 9, 3, 4, 6, 10, 2, 8, 5, 14, 12, 11, 15, 1,
????????????13, 6, 4, 9, 8, 15, 3, 0, 11, 1, 2, 12, 5, 10, 14, 7,
????????????1, 10, 13, 0, 6, 9, 8, 7, 4, 15, 14, 3, 11, 5, 2, 12
????????],
????
????????[
????????????7, 13, 14, 3, 0, 6, 9, 10, 1, 2, 8, 5, 11, 12, 4, 15,
????????????13, 8, 11, 5, 6, 15, 0, 3, 4, 7, 2, 12, 1, 10, 14, 9,
????????????10, 6, 9, 0, 12, 11, 7, 13, 15, 1, 3, 14, 5, 2, 8, 4,
????????????3, 15, 0, 6, 10, 1, 13, 8, 9, 4, 5, 11, 12, 7, 2, 14
????????],
????
????????[
????????????2, 12, 4, 1, 7, 10, 11, 6, 8, 5, 3, 15, 13, 0, 14, 9,
????????????14, 11, 2, 12, 4, 7, 13, 1, 5, 0, 15, 10, 3, 9, 8, 6,
????????????4, 2, 1, 11, 10, 13, 7, 8, 15, 9, 12, 5, 6, 3, 0, 14,
????????????11, 8, 12, 7, 1, 14, 2, 13, 6, 15, 0, 9, 10, 4, 5, 3
????????],
????
????????[
????????????12, 1, 10, 15, 9, 2, 6, 8, 0, 13, 3, 4, 14, 7, 5, 11,
????????????10, 15, 4, 2, 7, 12, 9, 5, 6, 1, 13, 14, 0, 11, 3, 8,
????????????9, 14, 15, 5, 2, 8, 12, 3, 7, 0, 4, 10, 1, 13, 11, 6,
????????????4, 3, 2, 12, 9, 5, 15, 10, 11, 14, 1, 7, 6, 0, 8, 13
????????],
????
????????[
????????????4, 11, 2, 14, 15, 0, 8, 13, 3, 12, 9, 7, 5, 10, 6, 1,
????????????13, 0, 11, 7, 4, 9, 1, 10, 14, 3, 5, 12, 2, 15, 8, 6,
????????????1, 4, 11, 13, 12, 3, 7, 14, 10, 15, 6, 8, 0, 5, 9, 2,
????????????6, 11, 13, 8, 1, 4, 10, 7, 9, 5, 0, 15, 14, 2, 3, 12
????????],
????
????????[
????????????13, 2, 8, 4, 6, 15, 11, 1, 10, 9, 13, 14, 5, 0, 12, 7,
????????????1, 15, 13, 8, 10, 3, 7, 4, 12, 5, 6, 11, 0, 14, 9, 2,
????????????7, 11, 4, 1, 9, 12, 14, 2, 0, 6, 10, 13, 15, 3, 5, 8,
????????????2, 1, 14, 7, 4, 10, 8, 13, 15, 12, 9, 0, 3, 5, 6, 11
????????]] ;
????????
????????public function DES()
????????{
????????????
????????}
????????
????????public static function getInstance():DES
????????{
????????????if (_instance == null)
????????????????_instance = new DES();
????????????return _instance;
????????}
????????
????????
????????
????}
}然后声明几个变量，用来存储有关内容，
复制内容到剪贴板代码:private var _bKey:ByteArray;????????????????????????//原始KEY (64 bit)
private var _bSubKey:ByteArray;????????????????????????//新的子KEY????( 16 * 48 bit)
private var _bOrigMsg:ByteArray;????????????????????//初始明文
private var _lMsg:ByteArray;????????????????????????//保存(L0-L16)
private var _rMsg:ByteArray;????????????????????????//保存R0-R16
private var _bCryptedMsg:ByteArray;????????????????//保存密文在DES的构造函数中初始化这些变量：
复制内容到剪贴板代码:_bKey = new ByteArray();
_bKey.length = 8;
_bSubKey = new ByteArray();
_bSubKey.length = 96;
????????????
_bOrigMsg = new ByteArray();
_bOrigMsg.length = 8;
????????????
_bCryptedMsg = new ByteArray();
_bCryptedMsg.length = 8;
????????????
_lMsg = new ByteArray();
_rMsg = new ByteArray();
_lMsg.length = _rMsg.length = 68;ByteArray是以字节位单位存储内容的,所以我们存储64位的数据块只需要长度为8的ByteArray对象,_bSubKey中我们要存储K1-K16的内容 ，每一个子密钥的位数是48位，6字节.因此需要6 * 16的长度。 _lMsg,_rMsg同理存储L0-L16，R0-R16 L为32位4字节,因此需要4 * 17 = 68字节长度来存储。 因为加密算法中都是以位为单位操作的，AS中没有直接操作位的语法(其实绝大多数语言都没有) 因此我们需要按位运算符来帮忙了。
这些东西对系统学过计算机的同学来说很容易了，如果是半路出家的可能平时没怎么接触过这些。因此我说得稍微啰嗦一点。
var b:ByteArray = new ByteArray();
b.writeByte(0);
这时b的内容用2进制表示就是:00000000 要想把某一位设为1需要使用 |(按位或运算符)，它对操作数中相应的位进行或运算。如果两个对应的位中有一个是1，结果位就是1。如果两个位都是0，结果就是0
00000000 | 10000000 = 01000000
也就是 b | 0x80 = 10000000
同理 b | 0x40 = 01000000
? ?? ? b | 0x20 = 00100000
? ?? ? b | 0x10 = 00010000
? ?? ? b | 0x8 = 00001000
? ?? ? b | 0x4 = 00000100
? ?? ? b | 0x2 = 00000010
? ?? ? b | 0x1 = 00000001
再说一个小技巧 16进制与2进制的转换，以前我的老师把它称为8421码，一个2进制数转换为16进制 每4位转换为一个16进制数 从高到底第一位是如果该位上是1那么数值就是8 第2位是4 第3位是2 第4位是1， 把他们相加就是16进制的表示, 如2进制数01001010 每4位一组 0100 1010
0 + 4 + 0 + 0 = 4 ；8 + 0 + 2 +０ ＝ １０(A) 16进制数就是4A。
如果要把某位上设为0需要使用& (按位与运算符) 它对操作数中相应的位进行与运算。如果相应的位都是1，结果位就是1，否则就是0。
设b = 11101000 , b & 0111111 = 0x01101000, 这个01111111怎么来的呢 把10000000取反就行了,10000000 16进制是0x80 ,那么~0x80 = 01111111; ~是取反运算符,可以反转操作数中的位，即1变成0，0变成1。
为了方便后面的操作 我们把0x80,0x40等等定义成DES类的一个属性数组：
//方便位操作的掩码
public static const BITMASK:Array = [0x01, 0x02, 0x04, 0x08, 0x10, 0x20, 0x40, 0x80 ];
定义setter方法来添加密钥与明文
????????/**
???????? * 设置初始密钥
???????? */
????????public function set bKey(value:ByteArray):void
????????{
????????????if (value.length < 8)
????????????{
????????????????throw new Error("密钥必须为64位");
????????????????return;
????????????}
????????????for (var i:int =??0; i < 8; i++ )
????????????{
????????????????_bKey[i] = value[i];
????????????}
????????}
????????
????????/**
???????? * 设置初始明文
???????? */
????????public function set bOrigMsg(value:ByteArray):void
????????{
????????????if (value.length < 8)
????????????{
????????????????throw new Error("明文必须为64位");
????????????????return;
????????????}
????????????
????????????for (var i:int =??0; i < 8; i++ )
????????????{
????????????????_bOrigMsg[i] = value[i];
????????????}
????????}如果输入的密钥长度小于64位则抛出错误，否则把密钥的内容复制到_bKey中。
定义一个辅助方法byteArraySet(byte, value:int),用来把byteArray对象的内容设为value
/**
???????? * 把ByteArray对象的内容设为指定值
???????? * @param????value
???????? */
????????public function byteArraySet(byteArray:ByteArray, value:int):void
????????{
????????????byteArray.position = 0;
????????????for (var i:int = 0, len:int = byteArray.length; i < len; i++ )
????????????{
????????????????byteArray.writeByte(value);
????????????}
????????}定义一个genSubKey方法用来生成子密钥: public function genSubKey():void {};
首先我们把_bSubKey的内容清空,
byteArraySet(_bSubKey, 0);
然后对初始密钥使用PC-1表进行置换，定义一个bTemp的byteArray对象来存储置换后的内容，
置换思路是这样的：依次检查_bKey中的PC-1表中索引的位，如果该位为1，则把bTemp中当前次数的位设为1，否则不做任何操作（因为我们初始化bTemp的时候，bTemp内容全部为0）。那么如何检查_bKey中指定的位是否为1呢？我们使用&（按位与运算符），这是按位与运算符，它对操作数中相应的位进行与运算。如果相应的位都是1，结果位就是1，否则就是0。
如果我们要检查0010 0001块中第3位是否为1,那么0010 0001 & 0010 0000结果为0010 0000，结果不为0表示0010 0001第3位是1,否则就是0。
这样我们需要检查_bKey中第10位是否为1那么 _bKey[int((10-1)/8)] & 0100 0000就是结果了(bKey中的元素是以字节为单位的，第10位那就处在第2个字节中的第2位，也就是_bKey[1]中的第2位,那么_bKey[1] & 0x40的结果如果不为0那么_bkey的第10位就是1，否则就是0。
var byteIndex:int, bitIndex:int, bTemp:ByteArray;
bTemp = new ByteArray();
bTemp.length = 7;
//置换选择1,对初始密钥使用PC-1置换
for (var i:int = 0; i < 56; i++ )
????????????{
? ?? ?? ?? ???
????????????????byteIndex = (bSelSwapTable_1[i] - 1) / 8;
????????????????bitIndex = 7 - (bSelSwapTable_1[i] - 1) % 8;
????????????????
????????????????if (_bKey[byteIndex] & BITMASK[bitIndex])
????????????????{
????????????????????bTemp[int(i / 8)] |= BITMASK[7 - i % 8];
????????????????}
????????????}byteIndex = (bSelSwapTable_1[i] - 1) / 8; 取得当前要检查的位在_bkey中的索引位置。
bitIndex = 7 - (bSelSwapTable_1[i] - 1) % 8; 取得当前要检查的为在_bkey中某索引的第几个位置
if (_bKey[byteIndex] & BITMASK[bitIndex])
如果当前位不为1，那么把bTemp中当前位设为1：
bTemp[int(i / 8)] |= BITMASK[7 - i % 8];
然后要经过16轮变换产生子密钥
先定一个bTempkey用来存储每轮变化时产生的值
var tmpLen:int = 0;
????????????//经过16轮变换产生子密钥
????????????for (var iTurn:int = 1; iTurn <= 16; iTurn++ )
????????????{
????????????????var bShiftBit:int = 2;
????????????????if (iTurn == 1 || iTurn == 2 || iTurn == 9 || iTurn == 16)
????????????????{
????????????????????bShiftBit = 1;
????????????????}
????????????????
????????????????tmpLen = bShiftBit + 28;
????????????????
????????????????//循环移位
????????????????for (i = bShiftBit;??i < tmpLen; i++ )
????????????????{
????????????????????
????????????????????//生成C(n)
????????????????????if (bTemp[int(i % 28 / 8)] & BITMASK[7 - (i % 28) % 8])
????????????????????{
????????????????????????bTempKey[int((i - bShiftBit) / 8)] |= BITMASK[7 - (i - bShiftBit) % 8];
????????????????????}
????????????????????
????????????????????//生成D(n)
????????????????????if (bTemp[int((i % 28 + 28) / 8)] & BITMASK[7 - (i % 28 + 28) % 8])
????????????????????{
????????????????????????bTempKey[int((i + 28 - bShiftBit) / 8)] |= BITMASK[7 - (i + 28 - bShiftBit) % 8];
????????????????????}
????????????????????
????????????????}
????????????????
????????????????
????????????????
????????????????//置换2
????????????????for (i = 0; i < 48; i++ )
????????????????{
????????????????????byteIndex = (bSelSwapTable_2[i] - 1) / 8;
????????????????????bitIndex = 7 - (bSelSwapTable_2[i] - 1) % 8;
????????????????????
????????????????????if (bTempKey[byteIndex] & BITMASK[bitIndex])
????????????????????{
????????????????????????_bSubKey[int(i / 8) + (iTurn - 1) * 6] |= BITMASK[7 - (i % 8)];
????????????????????}
????????????????}
????????????????byteArrayCpy(bTemp, bTempKey);
????????????????byteArraySet(bTempKey, 0);
????????????}首先按当前的轮数得到是左移一位还是两位
复制内容到剪贴板代码:var bShiftBit:int = 2;
????????????????if (iTurn == 1 || iTurn == 2 || iTurn == 9 || iTurn == 16)
????????????????{
????????????????????bShiftBit = 1;
????????????????}然后把bTemp分为Cn Dn(28位)进行循环，把结果存在bTempkey中,循环28次，如果移位数是1那么依次取出bTemp的2-3-4-5…28-1位的值设为bTempkey的1-28位的值， Dn就是Cn的基础上+上28就是了
复制内容到剪贴板代码:tmpLen = bShiftBit + 28;
????????????????
????????????????//循环移位
????????????????for (i = bShiftBit;??i < tmpLen; i++ )
????????????????{
????????????????????
????????????????????//生成C(n)
????????????????????if (bTemp[int(i % 28 / 8)] & BITMASK[7 - (i % 28) % 8])
????????????????????{
????????????????????????bTempKey[int((i - bShiftBit) / 8)] |= BITMASK[7 - (i - bShiftBit) % 8];
????????????????????}
????????????????????
????????????????????//生成D(n)
????????????????????if (bTemp[int((i % 28 + 28) / 8)] & BITMASK[7 - (i % 28 + 28) % 8])
????????????????????{
????????????????????????bTempKey[int((i + 28 - bShiftBit) / 8)] |= BITMASK[7 - (i + 28 - bShiftBit) % 8];
????????????????????}
????????????????????
????????????????}然后把CnDn用PC-2表进行置换 存储在_bSubKey中
复制内容到剪贴板代码://置换2
????????????????for (i = 0; i < 48; i++ )
????????????????{
????????????????????byteIndex = (bSelSwapTable_2[i] - 1) / 8;
????????????????????bitIndex = 7 - (bSelSwapTable_2[i] - 1) % 8;
????????????????????
????????????????????if (bTempKey[byteIndex] & BITMASK[bitIndex])
????????????????????{
????????????????????????_bSubKey[int(i / 8) + (iTurn - 1) * 6] |= BITMASK[7 - (i % 8)];
????????????????????}
????????????????}最后把bTempKey的值复制到bTemp中, 把bTemp的值清空,开始新一轮的循环
复制内容到剪贴板代码:byteArrayCpy(bTemp, bTempKey);
byteArraySet(bTempKey, 0);新建一个initSwap方法对明文进行初始置换
复制内容到剪贴板代码:/**
???????? * 根据IP表对明文M进行初始置换
???????? */
????????public function initSwap():void
????????{
????????????byteArraySet(_lMsg, 0);
????????????byteArraySet(_rMsg, 0);
????????????var byteIndex:int, bitIndex:int;
????????????for (var i:int = 0; i < 32; i++ )
????????????{
????????????????byteIndex = (bInitSwapTable[i] - 1) / 8;
????????????????bitIndex = 7 - (bInitSwapTable[i] - 1) % 8;
????????????????if (_bOrigMsg[byteIndex] & BITMASK[bitIndex])
????????????????{
????????????????????_lMsg[ int(i / 8)] |= BITMASK[7 - (i % 8)];
????????????????}
????????????????
????????????????byteIndex = (bInitSwapTable[i+32] - 1) / 8;
????????????????bitIndex = 7 - (bInitSwapTable[i + 32] - 1) % 8;
????????????????if (_bOrigMsg[byteIndex] & BITMASK[bitIndex])
????????????????{
????????????????????_rMsg[ int(i / 8)] |= BITMASK[7 - (i % 8)];
????????????????}
????????????}
????????}思路跟前面置换密钥是一样的，就不多说了.
新建一个singleTurn方法进行单轮转换
/**
???????? * 单轮加密
???????? * @param????iTurn
???????? * @param????mode
???????? */
????????public function singleTurn(iTurn:int, mode:Boolean):void
????????{
????????????//L(n) = R(n-1)
????????????for (var i:int = 0; i < 4; i++ )
????????????{
????????????????_lMsg[i + iTurn * 4] = _rMsg[i + (iTurn - 1) * 4];
????????????}
????????????
????????????F(iTurn, mode);
????????????
????????????for (i = 0; i < 4; i++ )
????????????{
????????????????_rMsg[i + iTurn * 4] ^= _lMsg[i + (iTurn - 1) * 4];
????????????????
????????????}
????????????
????????????
????????}//L(n) = R(n-1)
? ? ? ?? ? ? ?? ? ? ?for (var i:int = 0; i < 4; i++ )
? ? ? ?? ? ? ?? ? ? ?{
? ? ? ?? ? ? ?? ? ? ?? ? ? ?_lMsg[i + iTurn * 4] = _rMsg[i + (iTurn - 1) * 4];
? ? ? ?? ? ? ?? ? ? ?}
? ? ? ?把Rn-1的值复制到Ln中(Ln存储在_Lmsg[n] 到_lmsg[n+4]中, Rn同样)；
然后使用F函数对R(n-1)与Kn进行处理.F函数稍后讲解
F(iTurn, mode);
最后确定Rn的值（与Ln-1进行XOR操作）
for (var i:int = 0; i < 4; i++ )
? ? ? ?? ? ? ?? ? ? ?{
? ? ? ?? ? ? ?? ? ? ?? ? ? ?_lMsg[i + iTurn * 4] = _rMsg[i + (iTurn - 1) * 4];
? ? ? ?? ? ? ?? ? ? ?}
DES算法的关键部分就是F函数的实现,F函数输入的是右部的32位,首先通过扩展置换表,把32位扩展为48位，使他与子密钥长度相同，然后与子密钥进行XOR操作，再经过S盒处理，最后经过尾置换输出.
S盒的作用是把48位变成通过非线性转换为32位输出，总用有8个S盒，每一个S盒的作用是把6位变成4位输出，其中第1位和第6位组合作为行索引值，中间4位组合作为列索引值，例如对S盒输入的是101110，于是行索引=10（2进制）=2,列索引=0111(2进制) = 7， S1(2,7)=11 = 1011(2进制)，经过S盒之后还有一个尾置换.
/**
???????? * F函数
???????? * @param????iTurn????转换轮数
???????? * @param????mode????方式(加密OR解密)
???????? */
????????public function F(iTurn:int , mode:Boolean):void
????????{
????????????var bTemp:ByteArray = new ByteArray();
????????????bTemp.length = 6;
????????????var C:ByteArray = new ByteArray();
????????????C.length = 4;
????????????
????????????var byteIndex:int, bitIndex:int;
????????????
????????????//R --> E(R) 保存在bTemp中
????????????for (var i:int = 0; i < 48; i++ )
????????????{
????????????????byteIndex = (bBitExternTable[i] - 1) / 8;
????????????????bitIndex = 7 - (bBitExternTable[i] - 1) % 8;
????????????
????????????????if (_rMsg[byteIndex + (iTurn - 1) * 4] & BITMASK[bitIndex])
????????????????{
????????????????????bTemp[int(i / 8)] |= BITMASK[7 - i % 8];
????????????????}
????????????}
????
????????????//E(R) --> K 保存在bTemp中
????????????for (i = 0; i < 6; i++ )
????????????{
????????????????if (mode)
????????????????{
????????????????????bTemp[i] ^= _bSubKey[(iTurn - 1) * 6 + i];
????????????????}
????????????????else
????????????????{
????????????????????bTemp[i] ^= _bSubKey[96 - iTurn * 6 + i];
????????????????}
????????????}
????????
????????????var bStartPos:int = 0, bTarPos:int = 0;
????? ?? ???var bRowIndex:int = 0, bColIndex:int = 0 ;????//保存行数与列数
????????????for (var bBoxIndex:int = 0; bBoxIndex < 8; bBoxIndex ++ )
????????????{
????????????????// 确定行值
????????????????if ( bTemp[int(bStartPos / 8)] & BITMASK[(~bStartPos) & 7] )
????????????????{
????????????????????bColIndex += 2 ;
????????????????}
????????????????
????????????????if ( bTemp[int((bStartPos + 5) / 8)] & BITMASK[(~(bStartPos + 5)) & 7] )
????????????????{
????????????????????bColIndex += 1 ;
????????????????}
????????????????
????????????????// 确定列值
????????????????if ( bTemp[int((bStartPos+1)/8)] & BITMASK[(~(bStartPos+1))&7] )
????????????????????bRowIndex += 8 ;
????????????????if ( bTemp[int((bStartPos+2)/8)] & BITMASK[(~(bStartPos+2))&7] )
????????????????????bRowIndex += 4 ;
????????????????if ( bTemp[int((bStartPos+3)/8)] & BITMASK[(~(bStartPos+3))&7] )
????????????????????bRowIndex += 2 ;
????????????????if ( bTemp[int((bStartPos+4)/8)] & BITMASK[(~(bStartPos+4))&7] )
????????????????????bRowIndex += 1 ;
????????????????????
????????????????// 取出S(i)盒中由上面行列所对应的值bTarValue
????????????????var bTarValue:int = SB[bBoxIndex][bColIndex * 16 + bRowIndex];
????????????
????????????????
????????????????// 把bTarValue转化成对应的C(i)
????????????????if ( bTarValue >= 8 )
????????????????{
????????????????????C[int(bTarPos/8)] |= BITMASK[(~bTarPos)&7] ;
????????????????????bTarValue -= 8 ;
????????????????}
????????????????if ( bTarValue >= 4 )
????????????????{
????????????????????C[int((bTarPos+1)/8)] |= BITMASK[(~(bTarPos+1))&7] ;
????????????????????bTarValue -= 4 ;
????????????????}
????????????????if ( bTarValue >= 2 )
????????????????{
????????????????????C[int((bTarPos+2)/8)] |= BITMASK[(~(bTarPos+2))&7] ;
????????????????????bTarValue -= 2 ;
????????????????}
????????????????if ( bTarValue >= 1 )
????????????????{
????????????????????C[int((bTarPos+3)/8)] |= BITMASK[(~(bTarPos+3))&7] ;
????????????????????bTarValue -= 1 ;
????????????????}
????????????????
????????????????bRowIndex????= bColIndex = 0 ;
????????????????bStartPos????+= 6 ;
????????????????bTarPos????????+= 4 ;
????????????}
????????????
????????????// 尾置换
????????????for ( i = 0; i < 32; i++ )
????????????{
????????????????if ( C[int((bTailSwapTable[i]-1)/8)] & BITMASK[(~(bTailSwapTable[i]-1))&7] )
????????????????????_rMsg[iTurn *??4 + int(i /8)] |= BITMASK[(~i)&7] ;
????????????????else
????????????????????_rMsg[iTurn *??4 + int(i /8)] &= ( ~(BITMASK[(~i)&7]) );
????????????}
????????}
????????
????????
????}注意到这里常出现的BITMASK[(~(i))&7], 它其实就是BITMASK[7 - (i % 8)];因为位操作要比加减乘除要快得多，你可以把前面的操作都这样改过来，还有一个可以优化的地方就是int(i / 8)
可以改为 i >> 3;
最后创建一个函数initReSwap进行逆置换，生成密文复制内容到剪贴板代码:/**
???????? * 初始逆置换
???????? */
????????public function initReSwap():void
????????{
????????????byteArraySet(_bCryptedMsg, 0);
????????
????????????for ( var i:int = 0; i < 64; i++ )
????????????{
????????????????
????????????????if (bInitReSwapTable[i] <= 32)
????????????????{
????????????????????if (_rMsg[64 + int((bInitReSwapTable[i] - 1) / 8)] &
????????????????????? ? BITMASK[(~(bInitReSwapTable[i] - 1)) & 7])
????????????????????{
????????????????????????_bCryptedMsg[int(i / 8)] |= BITMASK[(~i) & 7];
????????????????????}
????????????????????else
????????????????????{
????????????????????????_bCryptedMsg[int(i / 8)] &= ( ~(BITMASK[(~i)&7]) ) ;
????????????????????}
????????????????????
????????????????}
????????????????else
????????????????{
????????????????????if (_lMsg[64 + int((bInitReSwapTable[i] - 1 - 32) / 8)] &
????????????????????? ? BITMASK[(~(bInitReSwapTable[i] - 32 - 1)) & 7])
????????????????????{
????????????????????????_bCryptedMsg[int(i / 8)] |= BITMASK[(~i) & 7];
????????????????????}
????????????????????else
????????????????????{
????????????????????????_bCryptedMsg[int(i / 8)] &= ( ~(BITMASK[(~i)&7]) ) ;
????????????????????}
????????????????}
????????????}
????????
????????????
????????}
????????
????}最终我们的DES加密类就完成了，记得给?_ bCryptedMsg设置一个getter
加密的时候对明文以64位进行分组，不足64的0补齐