我正在尝试为短字符串(短于64字节)实现简单的MD5。 我使用Wikipedia.的算法。 一切都在编译,但我的结果是字符串:
"hello world"
是:
BB3BB65ED0EE1EE0BB22CB93C3CD5A8F
虽然它应该是:
5EB63BBBE01EEED093CB22BB8F5ACDC3
完整的代码在这里:
program Prog;
uses Classes, SysUtils;
function leftrotate(x, c: Cardinal): Cardinal;
begin
leftrotate := (x shl c) or (x shr (32-c));
end;
const s: array[0..63] of Cardinal = (
7, 12, 17, 22, 7, 12, 17, 22, 7, 12, 17, 22, 7, 12, 17, 22,
5, 9, 14, 20, 5, 9, 14, 20, 5, 9, 14, 20, 5, 9, 14, 20,
4, 11, 16, 23, 4, 11, 16, 23, 4, 11, 16, 23, 4, 11, 16, 23,
6, 10, 15, 21, 6, 10, 15, 21, 6, 10, 15, 21, 6, 10, 15, 21 );
K: array[0..63] 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 );
var a0,b0,c0,d0, a,b,c,d, f,g,dTemp: Cardinal;
Len: Integer;
Msg: array[0..63] of Char;
M: array[0..15] of Cardinal absolute Msg; //break chunk into sixteen 32-bit words M[j]
Str: String;
i: Integer;
ff: TFileStream;
wait: Char;
begin
a0 := $67452301;
b0 := $efcdab89;
c0 := $98badcfe;
d0 := $10325476;
Str := 'hello world';
Len := Length(Str);
FillChar(Msg, 64, 0);
for i:=1 to Len do Msg[i-1] := Str[i];
//append "1" bit to message
Msg[Len] := chr(128);
//append original length in bits mod (2 pow 64) to message
Msg[63-7] := chr(8*Len); //Update thanks to @MBo
//Process each 512-bit chunk of message- 1 only have 1 chunk
//TEST dump
// ff := TFileStream.create('test.txt', fmCreate);
// ff.write(msg, 64);
// ff.free;
//Initialize hash value for this chunk:
A := a0;
B := b0;
C := c0;
D := d0;
//Main loop:
for i := 0 to 63 do begin
if (i>=0) and (i<=15) then begin
F := (B and C) or ((not B) and D);
g := i;
end
else if (i>=16) and (i<=31) then begin
F := (D and B) or ((not D) and C);
g := (5*i + 1) mod 16;
end
else if (i>=32) and (i<=47) then begin
F := B xor C xor D;
g := (3*i + 5) mod 16;
end
else if (i>=48) and (i<=63) then begin
F := C xor (B or (not D));
g := (7*i) mod 16;
end;
dTemp := D;
D := C;
C := B;
B := B + leftrotate((A + F + K[i] + M[g]), s[i]);
A := dTemp;
end;
//Add this chunk's hash to result so far:
a0 := a0 + A;
b0 := b0 + B;
c0 := c0 + C;
d0 := d0 + D;
//This should give 5EB63BBBE01EEED093CB22BB8F5ACDC3
Writeln( IntToHex(a0,8) + IntToHex(b0,8) + IntToHex(c0,8) +IntToHex(d0,8) );
Readln(wait);
end.
您可以在线试用代码: http://ideone.com/qdYQ6q
这是在主循环(test.txt)之前转储我准备好的块:
答案 0 :(得分:5)
最后一步是错误的:
a0 := a0 + A;
b0 := b0 + B;
c0 := c0 + C;
d0 := d0 + D;
它应该改变endianess:
a0 := Swap32(a0 + A);
b0 := Swap32(b0 + B);
c0 := Swap32(c0 + C);
d0 := Swap32(d0 + D);
function Swap32(ALong: Cardinal): Cardinal; Assembler;
asm
BSWAP eax
end;
然后很好。
答案 1 :(得分:3)
您可能需要考虑使用第三方实现,而不是创建自己的实现。例如,Indy的TIdHashMessageDigest5类生成正确的值,例如:
uses
..., IdHashMessageDigest;
var
S: string;
begin
with TIdHashMessageDigest5.Create do
try
S := HashStringAsHex('hello world'); // returns '5EB63BBBE01EEED093CB22BB8F5ACDC3'
finally
Free;
end;
end;
答案 2 :(得分:2)
这些步骤如何:
append "0" bit until message length in bits ≡ 448 (mod 512)
(56字节,64 + 56等) 和
append original length
位 mod (2 pow 64) to message
但你附加Len 字节
P.S。 我用Delphi检查了你的最后一个变种。我已将char类型更改为AnsiChar,结果与预期的一致。 请注意,正确的二进制结果不需要字节交换。它可能仅对从Int32值
构造十六进制字符串有所帮助 Int32在英特尔硬件上已经是小端,因此BB3BB65E
(十六进制字符串表示)对应于字节序列5E B6 3B BB
,依此类推。