关于md5的原理和算法的实现
谁能用比较通俗的方式讲解一下md5的实现原理。
另外如果要用C语言来实现一个md5算法,要怎么来实现,每个步骤都做了哪些工作。
733733
12 years, 4 months ago
Answers
来自维基百科:
//Note: All variables are unsigned 32 bit and wrap modulo 2^32 when calculating
var int[64] r, k
//r specifies the per-round shift amounts
r[ 0..15] := {7, 12, 17, 22, 7, 12, 17, 22, 7, 12, 17, 22, 7, 12, 17, 22}
r[16..31] := {5, 9, 14, 20, 5, 9, 14, 20, 5, 9, 14, 20, 5, 9, 14, 20}
r[32..47] := {4, 11, 16, 23, 4, 11, 16, 23, 4, 11, 16, 23, 4, 11, 16, 23}
r[48..63] := {6, 10, 15, 21, 6, 10, 15, 21, 6, 10, 15, 21, 6, 10, 15, 21}
//Use binary integer part of the sines of integers (Radians) as constants:
for i from 0 to 63
k[i] := floor(abs(sin(i + 1)) × (2 pow 32))
end for
//(Or just use the following table):
k[ 0.. 3] := { 0xd76aa478, 0xe8c7b756, 0x242070db, 0xc1bdceee }
k[ 4.. 7] := { 0xf57c0faf, 0x4787c62a, 0xa8304613, 0xfd469501 }
k[ 8..11] := { 0x698098d8, 0x8b44f7af, 0xffff5bb1, 0x895cd7be }
k[12..15] := { 0x6b901122, 0xfd987193, 0xa679438e, 0x49b40821 }
k[16..19] := { 0xf61e2562, 0xc040b340, 0x265e5a51, 0xe9b6c7aa }
k[20..23] := { 0xd62f105d, 0x02441453, 0xd8a1e681, 0xe7d3fbc8 }
k[24..27] := { 0x21e1cde6, 0xc33707d6, 0xf4d50d87, 0x455a14ed }
k[28..31] := { 0xa9e3e905, 0xfcefa3f8, 0x676f02d9, 0x8d2a4c8a }
k[32..35] := { 0xfffa3942, 0x8771f681, 0x6d9d6122, 0xfde5380c }
k[36..39] := { 0xa4beea44, 0x4bdecfa9, 0xf6bb4b60, 0xbebfbc70 }
k[40..43] := { 0x289b7ec6, 0xeaa127fa, 0xd4ef3085, 0x04881d05 }
k[44..47] := { 0xd9d4d039, 0xe6db99e5, 0x1fa27cf8, 0xc4ac5665 }
k[48..51] := { 0xf4292244, 0x432aff97, 0xab9423a7, 0xfc93a039 }
k[52..55] := { 0x655b59c3, 0x8f0ccc92, 0xffeff47d, 0x85845dd1 }
k[56..59] := { 0x6fa87e4f, 0xfe2ce6e0, 0xa3014314, 0x4e0811a1 }
k[60..63] := { 0xf7537e82, 0xbd3af235, 0x2ad7d2bb, 0xeb86d391 }
//Initialize variables:
var int h0 := 0x67452301 //A
var int h1 := 0xefcdab89 //B
var int h2 := 0x98badcfe //C
var int h3 := 0x10325476 //D
//Pre-processing: adding a single 1 bit
append "1" bit to message
/* Notice: the input bytes are considered as bits strings,
where the first bit is the most significant bit of the byte.[37]
//Pre-processing: padding with zeros
append "0" bit until message length in bit ≡ 448 (mod 512)
append length mod (2 pow 64) to message
//Process the message in successive 512-bit chunks:
for each 512-bit chunk of message
break chunk into sixteen 32-bit words w[j], 0 ≤ j ≤ 15
//Initialize hash value for this chunk:
var int a := h0
var int b := h1
var int c := h2
var int d := h3
//Main loop:
for i from 0 to 63
if 0 ≤ i ≤ 15 then
f := (b and c) or ((not b) and d)
g := i
else if 16 ≤ i ≤ 31
f := (d and b) or ((not d) and c)
g := (5×i + 1) mod 16
else if 32 ≤ i ≤ 47
f := b xor c xor d
g := (3×i + 5) mod 16
else if 48 ≤ i ≤ 63
f := c xor (b or (not d))
g := (7×i) mod 16
temp := d
d := c
c := b
b := b + leftrotate((a + f + k[i] + w[g]) , r[i])
a := temp
end for
//Add this chunk's hash to result so far:
h0 := h0 + a
h1 := h1 + b
h2 := h2 + c
h3 := h3 + d
end for
var char digest[16] := h0 append h1 append h2 append h3 //(Output is in little-endian)
//leftrotate function definition
leftrotate (x, c)
return (x << c) binary or (x >> (32-c));
大致算法上就是不停地移位、逻辑操作,然后反复多次。从布尔代数的角度上来说相当于做了一个非常复杂的非线性操作,所以最后产生的是一个随机结果。
isod3
answered 12 years, 4 months ago