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sp; Next
C15(26) = C14(0)
D15(26) = D14(0)
C15(27) = C14(1)
D15(27) = D14(1)
''''组合C15,D15成C_D
For i = 0 To 27
C_D(i) = C15(i)
C_D(i + 28) = D15(i)
Next
''''PC_2转换,生成K15
For i = 0 To 47
K15(i) = C_D(PC_2(i))
Next
''''***************************************************K16
''''生成C16,D16
For i = 0 To 26
C16(i) = C15(i + Lsi(16))
D16(i) = D15(i + Lsi(16))
Next
C16(27) = C15(0)
D16(27) = D15(0)
''''组合C16,D16成C_D
For i = 0 To 27
C_D(i) = C16(i)
C_D(i + 28) = D16(i)
Next
''''PC_2转换,生成K16
For i = 0 To 47
K16(i) = C_D(PC_2(i))
Next
''''**************************************************************************************
''''加密过程
''''将明文位数扩展为8的倍数
If (UBound(sCode) + 1) Mod 8 > 0 Then ReDim Preserve sCode(((UBound(sCode) + 1) \ 8 + 1) * 8 - 1)
''''定义返回密文长度
ReDim bReturn(UBound(sCode))
For j = 0 To UBound(sCode) Step 8
CopyMemory tempCode(0), sCode(j), 8
For i = 0 To 7
BinCode(i * 8 + 0) = (tempCode(i) And &H80) \ &H80
BinCode(i * 8 + 1) = (tempCode(i) And &H40) \ &H40
BinCode(i * 8 + 2) = (tempCode(i) And &H20) \ &H20
BinCode(i * 8 + 3) = (tempCode(i) And &H10) \ &H10
BinCode(i * 8 + 4) = (tempCode(i) And &H8) \ &H8
BinCode(i * 8 + 5) = (tempCode(i) And &H4) \ &H4
BinCode(i * 8 + 6) = (tempCode(i) And &H2) \ &H2
BinCode(i * 8 + 7) = (tempCode(i) And &H1) \ &H1
Next
''''IP置换
For i = 0 To 63
CodeIP(i) = BinCode(IP(i))
Next
''''分段
For i = 0 To 31
R0(i) = CodeIP(i)
L0(i) = CodeIP(i + 32)
Next
''''进行第一次迭代
For i = 0 To 47
CodeE(i) = R0(E(i)) ''''经过E变换扩充，由32位变为48位
CodeE(i) = CodeE(i) Xor K16(i) ''''与K16按位作不进位加法运算
Next
''''分8组
For i = 0 To 5
CodeS1(i) = CodeE(i)
CodeS2(i) = CodeE(i + 6)
CodeS3(i) = CodeE(i + 12)
CodeS4(i) = CodeE(i + 18)
CodeS5(i) = CodeE(i + 24)
CodeS6(i) = CodeE(i + 30)
CodeS7(i) = CodeE(i + 36)
CodeS8(i) = CodeE(i + 42)
Next
''''S盒运算，得到8个数
S(0) = S1(CodeS1(5) + CodeS1(0) * 2, CodeS1(4) + CodeS1(3) * 2 + CodeS1(2) * 4 + CodeS1(1) * 8)
S(1) = S2(CodeS2(5) + CodeS2(0) * 2, CodeS2(4) + CodeS2(3) * 2 + CodeS2(2) * 4 + CodeS2(1) * 8)
S(2) = S3(CodeS3(5) + CodeS3(0) * 2, CodeS3(4) + CodeS3(3) * 2 + CodeS3(2) * 4 + CodeS3(1) * 8)
S(3) = S4(CodeS4(5) + CodeS4(0) * 2, CodeS4(4) + CodeS4(3) * 2 + CodeS4(2) * 4 + CodeS4(1) * 8)
S(4) = S5(CodeS5(5) + CodeS5(0) * 2, CodeS5(4) + CodeS5(3) * 2 + CodeS5(2) * 4 + CodeS5(1) * 8)
S(5) = S6(CodeS6(5) + CodeS6(0) * 2, CodeS6(4) + CodeS6(3) * 2 + CodeS6(2) * 4 + CodeS6(1) * 8)
S(6) = S7(CodeS7(5) + CodeS7(0) * 2, CodeS7(4) + CodeS7(3) * 2 + CodeS7(2) * 4 + CodeS7(1) * 8)
S(7) = S8(CodeS8(5) + CodeS8(0) * 2, CodeS8(4) + CodeS8(3) * 2 + CodeS8(2) * 4 + CodeS8(1) * 8)
''''S盒运算32位结果
For i = 0 To 7
RetS(i * 4 + 0) = (S(i) And &H8) \ &H8
RetS(i * 4 + 1) = (S(i) And &H4) \ &H4
RetS(i * 4 + 2) = (S(i) And &H2) \ &H2
RetS(i * 4 + 3) = (S(i) And &H1) \ &H1
Next
For i = 0 To 31
''''P变换
CodeP(i) = RetS(P(i))
''''产生L1,R1
R1(i) = L0(i) Xor CodeP(i)
L1(i) = R0(i)
Next
''''进行第二次迭代
For i = 0 To 47
CodeE(i) = R1(E(i)) ''''经过E变换扩充，由32位变为48位
CodeE(i) = CodeE(i) Xor K15(i) ''''与K15按位作不进位加法运算
Next
''''分8组
For i = 0 To 5
CodeS1(i) = CodeE(i)
CodeS2(i) = CodeE(i + 6)
CodeS3(i) = CodeE(i + 12)
CodeS4(i) = CodeE(i + 18)
CodeS5(i) = CodeE(i + 24)
CodeS6(i) = CodeE(i + 30)
CodeS7(i) = CodeE(i + 36)
CodeS8(i) = CodeE(i + 42)
Next
''''S盒运算，得到8个数
S(0) = S1(CodeS1(5) + CodeS1(0) * 2, CodeS1(4) + CodeS1(3) * 2 + CodeS1(2) * 4 + CodeS1(1) * 8)
S(1) = S2(CodeS2(5) + CodeS2(0) * 2, CodeS2(4) + CodeS2(3) * 2 + CodeS2(2) * 4 + CodeS2(1) * 8)
S(2) = S3(CodeS3(5) + CodeS3(0) * 2, CodeS3(4) + CodeS3(3) * 2 + CodeS3(2) * 4 + CodeS3(1) * 8)
S(3) = S4(CodeS4(5) + CodeS4(0) * 2, CodeS4(4) + CodeS4(3) * 2 + CodeS4(2) * 4 + CodeS4(1) * 8)
S(4) = S5(CodeS5(5) + CodeS5(0) * 2, CodeS5(4) + CodeS5(3) * 2 + CodeS5(2) * 4 + CodeS5(1) * 8)
S(5) = S6(CodeS6(5) + CodeS6(0) * 2, CodeS6(4) + CodeS6(3) * 2 + CodeS6(2) * 4 + CodeS6(1) * 8)
S(6) = S7(CodeS7(5) + CodeS7(0) * 2, CodeS7(4) + CodeS7(3) * 2 + CodeS7(2) * 4 + CodeS7(1) * 8)
S(7) = S8(CodeS8(5) + CodeS8(0) * 2, CodeS8(4) + CodeS8(3) * 2 + CodeS8(2) * 4 + CodeS8(1) * 8)
''''S盒运算32位结果
For i = 0 To 7
RetS(i * 4 + 0) = (S(i) And &H8) \ &H8
RetS(i * 4 + 1) = (S(i) And &H4) \ &H4
RetS(i * 4 + 2) = (S(i) And &H2) \ &H2
RetS(i * 4 + 3) = (S(i) And &H1) \ &H1
Next
For i = 0 To 31
''''P变换
CodeP(i) = RetS(P(i))
''''产生L2,R2
R2(i) = L1(i) Xor CodeP(i)
L2(i) = R1(i)
Next
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