对一个文件(任何类型的文件)用MD5算法加密，加密前是否要对文件做一些预处理？

主    题：用Delphi实现MD5算法
作       者:    BlueSong
等    级：版主
信    誉： 100
所属论坛：开发语言 >> Delphi
问题分数： 10
发表时间： 2003-04-11 14:03 --------------------------------------------------------------------------------

unit MD5;interface  procedure MD5Code(var InData; var OutData; cByte: Integer);implementationuses
  Windows, SysUtils;const
   Ca = $67452301;
   Cb = $EFCDAB89;
   Cc = $98BADCFE;
   Cd = $10325476;  T: array [0..3,0..15] of Cardinal =
   (($d76aa478,
     $e8c7b756,
     $242070db,
     $c1bdceee,
     $f57c0faf,
     $4787c62a,
     $a8304613,
     $fd469501,
     $698098d8,
     $8b44f7af,
     $ffff5bb1,
     $895cd7be,
     $6b901122,
     $fd987193,
     $a679438e,
     $49b40821),    ($f61e2562,
     $c040b340,
     $265e5a51,
     $e9b6c7aa,
     $d62f105d,
     $02441453,
     $d8a1e681,
     $e7d3fbc8,
     $21e1cde6,
     $c33707d6,
     $f4d50d87,
     $455a14ed,
     $a9e3e905,
     $fcefa3f8,
     $676f02d9,
     $8d2a4c8a),    ($fffa3942,
     $8771f681,
     $6d9d6122,
     $fde5380c,
     $a4beea44,
     $4bdecfa9,
     $f6bb4b60,
     $bebfbc70,
     $289b7ec6,
     $eaa127fa,
     $d4ef3085,
     $04881d05,
     $d9d4d039,
     $e6db99e5,
     $1fa27cf8,
     $c4ac5665),    ($f4292244,
     $432aff97,
     $ab9423a7,
     $fc93a039,
     $655b59c3,
     $8f0ccc92,
     $ffeff47d,
     $85845dd1,
     $6fa87e4f,
     $fe2ce6e0,
     $a3014314,
     $4e0811a1,
     $f7537e82,
     $bd3af235,
     $2ad7d2bb,
     $eb86d391));function ROL (Value: Cardinal; Shift: Byte): Cardinal;
begin
  Assert(Shift <= 32);
  Result := (Value shl Shift) or (Value shr (32 - Shift));
end;procedure FF(var a: Cardinal; b,c,d,Mj,s,ti: Cardinal);
  function F(X, Y, Z: Cardinal): Cardinal;
  begin
    Result :=(X and Y) or ((not X) and Z);
  end;
begin
  a := b +  ROL(a +(F(b,c,d) + Mj + ti), s);
end;procedure GG(var a: Cardinal; b,c,d,Mj,s,ti: Cardinal);
  function G(X, Y, Z: Cardinal): Cardinal;
  begin
    Result := (X and Z) or (Y and (not Z));
  end;
begin
  a := b +  ROL(a +(G(b,c,d) + Mj + ti), s);
end;procedure HH(var a: Cardinal; b,c,d,Mj,s,ti: Cardinal);
  function H(X, Y, Z: Cardinal): Cardinal;
  begin
    Result := X xor Y xor Z;
  end;
begin
  a := b +  ROL(a +(H(b,c,d) + Mj + ti), s);
end;procedure II(var a: Cardinal; b,c,d,Mj,s,ti: Cardinal);
  function I(X, Y, Z: Cardinal): Cardinal;
  begin
    Result := Y xor (X or (not Z));
  end;
begin
  a := b +  ROL(a +(I(b,c,d) + Mj + ti), s);
end;procedure MD5Code(var InData; var OutData; cByte: Integer);
type
  TDWordAry = array [0..3] of Cardinal;
  PDWordAry = ^TDwordAry;
var
  M: array[0..15] of Cardinal;
  CGroup: Integer;
  i: Integer;
  Va,Vb,Vc,Vd: Cardinal;
  a,b,c,d: Cardinal;
  OldLen: Integer;
begin
  OldLen := cByte;
  CGroup := (cByte + 1) div 64;
  if (cByte + 1) mod 64 > 56 then
    Inc(CGroup);  Va := Ca;
  Vb := Cb;
  Vc := Cc;
  Vd := Cd;  for i := 0 to CGroup do
  begin
    a := Va;
    b := Vb;
    c := Vc;
    d := Vd;    if i = CGroup then
    begin
      ZeroMemory(@M[0], 64);
      if cByte >= 0 then
      begin
        Move(PByteArray(@InData)^[i * 64], M[0], cByte);
        PByteArray(@M)^[cByte] := $80;
      end;
      M[14] := OldLen shl 3;
      M[15] := OldLen shr 29;
    end
    else
    begin
      if cByte < 64 then
      begin
        Move(PByteArray(@InData)^[i * 64], M[0], cByte);
        PByteArray(@M)^[cByte] := $80;
        ZeroMemory(@PByteArray(@M)^[cByte + 1], 64 - cByte - 1);
      end
      else
        Move(PByteArray(@InData)^[i * 64], M[0], 64);
      Dec(cByte, 64);
    end;
      //第一轮
     FF(a,b,c,d,M[0],7,  T[0,0]);
     FF(d,a,b,c,M[1],12, T[0,1]);
     FF(c,d,a,b,M[2],17, T[0,2]);
     FF(b,c,d,a,M[3],22, T[0,3]);
     FF(a,b,c,d,M[4],7,  T[0,4]);
     FF(d,a,b,c,M[5],12, T[0,5]);
     FF(c,d,a,b,M[6],17, T[0,6]);
     FF(b,c,d,a,M[7],22, T[0,7]);
     FF(a,b,c,d,M[8],7,  T[0,8]);
     FF(d,a,b,c,M[9],12, T[0,9]);
     FF(c,d,a,b,M[10],17,T[0,10]);
     FF(b,c,d,a,M[11],22,T[0,11]);
     FF(a,b,c,d,M[12],7, T[0,12]);
     FF(d,a,b,c,M[13],12,T[0,13]);
     FF(c,d,a,b,M[14],17,T[0,14]);
     FF(b,c,d,a,M[15],22,T[0,15]);
     //第二轮
     GG(a,b,c,d,M[1],5,   T[1,0]);
     GG(d,a,b,c,M[6],9,   T[1,1]);
     GG(c,d,a,b,M[11],14, T[1,2]);
     GG(b,c,d,a,M[0],20,  T[1,3]);
     GG(a,b,c,d,M[5],5,   T[1,4]);
     GG(d,a,b,c,M[10],9,  T[1,5]);
     GG(c,d,a,b,M[15],14, T[1,6]);
     GG(b,c,d,a,M[4],20,  T[1,7]);
     GG(a,b,c,d,M[9],5,   T[1,8]);
     GG(d,a,b,c,M[14],9,  T[1,9]);
     GG(c,d,a,b,M[3],14,  T[1,10]);
     GG(b,c,d,a,M[8],20,  T[1,11]);
     GG(a,b,c,d,M[13],5,  T[1,12]);
     GG(d,a,b,c,M[2],9,   T[1,13]);
     GG(c,d,a,b,M[7],14,  T[1,14]);
     GG(b,c,d,a,M[12],20, T[1,15]);
     //第三轮
     HH(a,b,c,d,M[5],4,   T[2,0]);
     HH(d,a,b,c,M[8],11,  T[2,1]);
     HH(c,d,a,b,M[11],16, T[2,2]);
     HH(b,c,d,a,M[14],23, T[2,3]);
     HH(a,b,c,d,M[1],4,   T[2,4]);
     HH(d,a,b,c,M[4],11,  T[2,5]);
     HH(c,d,a,b,M[7],16,  T[2,6]);
     HH(b,c,d,a,M[10],23, T[2,7]);
     HH(a,b,c,d,M[13],4,  T[2,8]);
     HH(d,a,b,c,M[0],11,  T[2,9]);
     HH(c,d,a,b,M[3],16,  T[2,10]);
     HH(b,c,d,a,M[6],23,  T[2,11]);
     HH(a,b,c,d,M[9],4,   T[2,12]);
     HH(d,a,b,c,M[12],11, T[2,13]);
     HH(c,d,a,b,M[15],16, T[2,14]);
     HH(b,c,d,a,M[2],23,  T[2,15]);
     //第四轮
     II(a,b,c,d,M[0],6,   T[3,0]);
     II(d,a,b,c,M[7],10,  T[3,1]);
     II(c,d,a,b,M[14],15, T[3,2]);
     II(b,c,d,a,M[5],21,  T[3,3]);
     II(a,b,c,d,M[12],6,  T[3,4]);
     II(d,a,b,c,M[3],10,  T[3,5]);
     II(c,d,a,b,M[10],15, T[3,6]);
     II(b,c,d,a,M[1],21,  T[3,7]);
     II(a,b,c,d,M[8],6,   T[3,8]);
     II(d,a,b,c,M[15],10, T[3,9]);
     II(c,d,a,b,M[6],15,  T[3,10]);
     II(b,c,d,a,M[13],21, T[3,11]);
     II(a,b,c,d,M[4],6,   T[3,12]);
     II(d,a,b,c,M[11],10, T[3,13]);
     II(c,d,a,b,M[2],15,  T[3,14]);
     II(b,c,d,a,M[9],21,  T[3,15]);    Va := Va + a;
    Vb := Vb + b;
    Vc := Vc + c;
    Vd := Vd + d;
  end;  PDWordAry(@OutData)^[0] := Va;
  PDWordAry(@OutData)^[1] := Vb;
  PDWordAry(@OutData)^[2] := Vc;
  PDwordAry(@OutData)^[3] := Vd;
end;
end.

(*********************from:http://www.freesoft.org/CIE/RFC/1321/3.htm***********MD5 Algorithm Description  We begin by supposing that we have a b-bit message as input, and that we wish
to find its message digest. Here b is an arbitrary nonnegative integer; b may be
zero, it need not be a multiple of eight, and it may be arbitrarily large. We
imagine the bits of the message written down as follows:
          m_0 m_1 ... m_{b-1}  The following five steps are performed to compute the message digest of the
message.Step 1. Append Padding Bits  The message is "padded" (extended) so that its length (in bits) is congruent
to 448, modulo 512. That is, the message is extended so that it is just 64 bits
shy of being a multiple of 512 bits long. Padding is always performed, even if
the length of the message is already congruent to 448, modulo 512.  Padding is performed as follows: a single "1" bit is appended to the message,
and then "0" bits are appended so that the length in bits of the padded message
becomes congruent to 448, modulo 512. In all, at least one bit and at most 512
bits are appended.Step 2. Append Length  A 64-bit representation of b (the length of the message before the padding
bits were added) is appended to the result of the previous step. In the unlikely
event that b is greater than 2^64, then only the low-order 64 bits of b are
used. (These bits are appended as two 32-bit words and appended low-order word
first in accordance with the previous conventions.)  At this point the resulting message (after padding with bits and with b) has
a length that is an exact multiple of 512 bits. Equivalently, this message has
a length that is an exact multiple of 16 (32-bit) words. Let M[0 ... N-1] denote
the words of the resulting message, where N is a multiple of 16.Step 3. Initialize MD Buffer  A four-word buffer (A,B,C,D) is used to compute the message digest. Here each
of A, B, C, D is a 32-bit register. These registers are initialized to the
following values in hexadecimal, low-order bytes first):
          word A: 01 23 45 67
          word B: 89 ab cd ef
          word C: fe dc ba 98
          word D: 76 54 32 10Step 4. Process Message in 16-Word Blocks  We first define four auxiliary functions that each take as input three 32-bit
words and produce as output one 32-bit word.
　　　F(X,Y,Z) =(X&Y)|((~X)&Z)
　　　G(X,Y,Z) =(X&Z)|(Y&(~Z))
　　　H(X,Y,Z) =X^Y^Z
　　　I(X,Y,Z)=Y^(X|(~Z))  In each bit position F acts as a conditional: if X then Y else Z. The function
F could have been defined using + instead of v since XY and not(X)Z will never
have 1's in the same bit position.) It is interesting to note that if the bits
of X, Y, and Z are independent and unbiased, the each bit of F(X,Y,Z) will be
independent and unbiased.  The functions G, H, and I are similar to the function F, in that they act in
"bitwise parallel" to produce their output from the bits of X, Y, and Z, in such
a manner that if the corresponding bits of X, Y, and Z are independent and
unbiased, then each bit of G(X,Y,Z), H(X,Y,Z), and I(X,Y,Z) will be independent
and unbiased. Note that the function H is the bit-wise "xor" or "parity"
function of its inputs.  This step uses a 64-element table T[1 ... 64] constructed from the sine
function. Let T[i] denote the i-th element of the table, which is equal to the
integer part of 4294967296 times abs(sin(i)), where i is in radians. The
elements of the table are given in the appendix.Do the following:
   /* Process each 16-word block. */
   For i = 0 to N/16-1 do
     /* Copy block i into X. */
     For j = 0 to 15 do
       Set X[j] to M[i*16+j].
     end /* of loop on j */     /* Save A as AA, B as BB, C as CC, and D as DD. */
     AA = A
     BB = B     CC = C
     DD = D     /* Round 1. */
     /* Let [abcd k s i] denote the operation
          a = b + ((a + F(b,c,d) + X[k] + T[i]) <<< s). */
     /* Do the following 16 operations. */
     [ABCD  0  7  1]
     [DABC  1 12  2]
     [CDAB  2 17  3]
     [BCDA  3 22  4]
     [ABCD  4  7  5]
     [DABC  5 12  6]
     [CDAB  6 17  7]
     [BCDA  7 22  8]
     [ABCD  8  7  9]
     [DABC  9 12 10]
     [CDAB 10 17 11]
     [BCDA 11 22 12]
     [ABCD 12  7 13]
     [DABC 13 12 14]
     [CDAB 14 17 15]
     [BCDA 15 22 16]     /* Round 2. */
     /* Let [abcd k s i] denote the operation
          a = b + ((a + G(b,c,d) + X[k] + T[i]) <<< s). */
     /* Do the following 16 operations. */
     [ABCD  1  5 17]
     [DABC  6  9 18]
     [CDAB 11 14 19]
     [BCDA  0 20 20]
     [ABCD  5  5 21]
     [DABC 10  9 22]
     [CDAB 15 14 23]
     [BCDA  4 20 24]
     [ABCD  9  5 25]
     [DABC 14  9 26]
     [CDAB  3 14 27]
     [BCDA  8 20 28]
     [ABCD 13  5 29]
     [DABC  2  9 30]
     [CDAB  7 14 31]
     [BCDA 12 20 32]     /* Round 3. */
     /* Let [abcd k s t] denote the operation
          a = b + ((a + H(b,c,d) + X[k] + T[i]) <<< s). */
     /* Do the following 16 operations. */
     [ABCD  5  4 33]
     [DABC  8 11 34]
     [CDAB 11 16 35]
     [BCDA 14 23 36]
     [ABCD  1  4 37]
     [DABC  4 11 38]
     [CDAB  7 16 39]
     [BCDA 10 23 40]
     [ABCD 13  4 41]
     [DABC  0 11 42]
     [CDAB  3 16 43]
     [BCDA  6 23 44]
     [ABCD  9  4 45]
     [DABC 12 11 46]
     [CDAB 15 16 47]
     [BCDA  2 23 48]     /* Round 4. */
     /* Let [abcd k s t] denote the operation
          a = b + ((a + I(b,c,d) + X[k] + T[i]) <<< s). */
     /* Do the following 16 operations. */
     [ABCD  0  6 49]
     [DABC  7 10 50]
     [CDAB 14 15 51]
     [BCDA  5 21 52]
     [ABCD 12  6 53]
     [DABC  3 10 54]
     [CDAB 10 15 55]
     [BCDA  1 21 56]
     [ABCD  8  6 57]
     [DABC 15 10 58]
     [CDAB  6 15 59]
     [BCDA 13 21 60]
     [ABCD  4  6 61]
     [DABC 11 10 62]
     [CDAB  2 15 63]
     [BCDA  9 21 64]     /* Then perform the following additions. (That is increment each
        of the four registers by the value it had before this block
        was started.) */
     A = A + AA
     B = B + BB
     C = C + CC
     D = D + DD   end /* of loop on i */Step 5. Output  The message digest produced as output is A, B, C, D. That is, we begin with
the low-order byte of A, and end with the high-order byte of D.  This completes the description of MD5. A reference implementation in C is
given in the appendix.Summary  The MD5 message-digest algorithm is simple to implement, and provides a
"fingerprint" or message digest of a message of arbitrary length. It is
conjectured that the difficulty of coming up with two messages having the same
message digest is on the order of 2^64 operations, and that the difficulty of
coming up with any message having a given message digest is on the order of
2^128 operations. The MD5 algorithm has been carefully scrutinized for
weaknesses. It is, however, a relatively new algorithm and further security
analysis is of course justified, as is the case with any new proposal of this
sort.*******************************************************************************)

to  lovefox_zoe(爱情狐狸)：
谢谢你的代码，能否针对你代码的说明一下你的思路，我看起代码来有点费劲。
或者，给下面几句代码增加注释，谢谢！
//InData,OutData是什么类型？
procedure MD5Code(var InData; var OutData; cByte: Integer);//这四个初始值是不是和你下面写的内容不一致，为什么？
const
   Ca = $67452301;
   Cb = $EFCDAB89;
   Cc = $98BADCFE;
   Cd = $10325476;

我就结合一些网络资料翻译一下，MD5加密原理吧。
一、算法实现
1、MD5算法是对输入的数据进行补位，使得如果数据位长度LEN对512求余的结果
是448。
即数据扩展至K*512+448位。即K*64+56个字节，K为整数。
具体补位操作：补一个1，然后补0至满足上述要求
2、补数据长度：
用一个64位的数字表示数据的原始长度B，把B用两个32位数表示。这时，数据
就被填
补成长度为512位的倍数。
3. 初始化MD5参数
四个32位整数 (A,B,C,D) 用来计算信息摘要，初始化使用的是十六进制表示
的数字
A=0X01234567
B=0X89abcdef
C=0Xfedcba98
D=0X76543210
4、处理位操作函数
X，Y，Z为32位整数。
F(X,Y,Z) = X&Y|NOT(X)&Z
G(X,Y,Z) = X&Z|Y¬(Z)
H(X,Y,Z) = X xor Y xor Z
I(X,Y,Z) = Y xor (X|not(Z))
5、主要变换过程：
使用常数组T[1 ... 64]， T为32位整数用16进制表示，数据用16个32位的
整
数数组M[]表示。
具体过程如下：
/* 处理数据原文 */
For i = 0 to N/16-1 do
/*每一次，把数据原文存放在16个元素的数组X中. */
For j = 0 to 15 do
Set X[j] to M[i*16+j].
end /结束对J的循环
/* Save A as AA, B as BB, C as CC, and D as DD. */
AA = A
BB = B
CC = C
DD = D
/* 第1轮*/
/* 以 [abcd k s i]表示如下操作
a = b + ((a + F(b,c,d) + X[k] + T) <<< s). */
/* Do the following 16 operations. */
[ABCD 0 7 1] [DABC 1 12 2] [CDAB 2 17 3] [BCDA 3 22 4]
[ABCD 4 7 5] [DABC 5 12 6] [CDAB 6 17 7] [BCDA 7 22 8]
[ABCD 8 7 9] [DABC 9 12 10] [CDAB 10 17 11] [BCDA 11 22 12]
[ABCD 12 7 13] [DABC 13 12 14] [CDAB 14 17 15] [BCDA 15 22 16]
/* 第2轮* */
/* 以 [abcd k s i]表示如下操作
a = b + ((a + G(b,c,d) + X[k] + T) <<< s). */
/* Do the following 16 operations. */
[ABCD 1 5 17] [DABC 6 9 18] [CDAB 11 14 19] [BCDA 0 20 20]
[ABCD 5 5 21] [DABC 10 9 22] [CDAB 15 14 23] [BCDA 4 20 24]
[ABCD 9 5 25] [DABC 14 9 26] [CDAB 3 14 27] [BCDA 8 20 28]
[ABCD 13 5 29] [DABC 2 9 30] [CDAB 7 14 31] [BCDA 12 20 32]
/* 第3轮*/
/* 以 [abcd k s i]表示如下操作
a = b + ((a + H(b,c,d) + X[k] + T) <<< s). */
/* Do the following 16 operations. */
[ABCD 5 4 33] [DABC 8 11 34] [CDAB 11 16 35] [BCDA 14 23 36]
[ABCD 1 4 37] [DABC 4 11 38] [CDAB 7 16 39] [BCDA 10 23 40]
[ABCD 13 4 41] [DABC 0 11 42] [CDAB 3 16 43] [BCDA 6 23 44]
[ABCD 9 4 45] [DABC 12 11 46] [CDAB 15 16 47] [BCDA 2 23 48]
/* 第4轮*/
/* 以 [abcd k s i]表示如下操作
a = b + ((a + I(b,c,d) + X[k] + T) <<< s). */
/* Do the following 16 operations. */
[ABCD 0 6 49] [DABC 7 10 50] [CDAB 14 15 51] [BCDA 5 21 52]
[ABCD 12 6 53] [DABC 3 10 54] [CDAB 10 15 55] [BCDA 1 21 56]
[ABCD 8 6 57] [DABC 15 10 58] [CDAB 6 15 59] [BCDA 13 21 60]
[ABCD 4 6 61] [DABC 11 10 62] [CDAB 2 15 63] [BCDA 9 21 64]
/* 然后进行如下操作 */
A = A + AA
B = B + BB
C = C + CC
D = D + DD
end /* 结束对I的循环*/
6、输出结果。

to lovefox_zoe(爱情狐狸)：非常感谢你的回答，我麻烦你一下，给我一个通过使用MD5Code过程加密的例子，包括对字符串和文件加密，因为我对下面的过程中的InData和OutData的用法不了解。procedure MD5Code(var InData; var OutData; cByte: Integer);

to lovefox_zoe：你给我的代码由于没有注释，我看起来很不容易，能不能给下面的代码加上注释，否则我真是一头雾水。
OldLen := cByte;
  CGroup := (cByte + 1) div 64;     //为什么cByte加1
  if (cByte + 1) mod 64 > 56 then   //这是什么意思
    Inc(CGroup);
if i = CGroup then
    begin
      ZeroMemory(@M[0], 64);
      if cByte >= 0 then
      begin
        Move(PByteArray(@InData)^[i * 64], M[0], cByte);
        PByteArray(@M)^[cByte] := $80;
      end;
      M[14] := OldLen shl 3;      //这两句的目的是什么？
      M[15] := OldLen shr 29;
    end
    else
    begin
      if cByte < 64 then
      begin
        Move(PByteArray(@InData)^[i * 64], M[0], cByte);
        PByteArray(@M)^[cByte] := $80;
        ZeroMemory(@PByteArray(@M)^[cByte + 1], 64 - cByte - 1);
      end
      else
        Move(PByteArray(@InData)^[i * 64], M[0], 64);
      Dec(cByte, 64);
    end;谢谢！

调试易

对一个文件(任何类型的文件)用MD5算法加密，加密前是否要对文件做一些预处理？

解决方案 »