演算法c語言實現英文版

『壹』 C語言英文文本加密

#include "stdio.h"

#include <stdlib.h>

int main(int argc,char *argv[]){

FILE *fp,*fq;

int k,t;

fp=fopen("AAA12345678901.txt","w+");

if(!fp || (fq=fopen("tmp.txt","w"))==NULL){

printf("Failed to open the file and exit... ");

return 0;

}

printf("Please enter a short passage(letters+space+punctuation,'Enter' end)... ");

while((t=getchar())!=' ')//為文件輸入內容

fputc(t,fp);

printf("Please enter the encryption key(int >0)... k=");

while(scanf("%d",&k)!=1 || k<1){//輸入加密密鑰並判斷是否正確

printf("Input error, redo: ");

fflush(stdin);

}

rewind(fp);

while(t=fgetc(fp),!feof(fp))//加密

if(t>='A' && t<='Z')

fputc(((t-'A')+k)%26+'A',fq);

else if(t>='a' && t<='z')

fputc(((t-'a')+k)%26+'a',fq);

else

fputc(t,fq);

fclose(fp);//關閉原文件

fclose(fq);//關閉加密後的文件

remove("AAA12345678901.txt");//刪除原文件

rename("tmp.txt","AAA12345678901.txt");//將加密後的文件更換為原文件名

printf(" ");

if(fp=fopen("AAA12345678901.txt","r")){

while((t=fgetc(fp))!=EOF)

printf("%c",t);

printf(" Encryption success! ");

}

else

printf(" Failed to open the encrypted file... ");

fclose(fp);

return 0;

}

代碼格式和運行樣例圖片：

『貳』 如何用C語言實現RSA演算法

RSA演算法它是第一個既能用於數據加密也能用於數字簽名的演算法。它易於理解和操作，也很流行。演算法的名字以發明者的名字
命名：Ron Rivest, Adi Shamir 和Leonard
Adleman。但RSA的安全性一直未能得到理論上的證明。它經歷了各種攻擊，至今未被完全攻破。

一、RSA演算法 :

首先, 找出三個數, p, q, r,
其中 p, q 是兩個相異的質數, r 是與 (p-1)(q-1) 互質的數
p, q, r 這三個數便是 private key

接著, 找出 m, 使得 rm == 1 mod (p-1)(q-1)
這個 m 一定存在, 因為 r 與 (p-1)(q-1) 互質, 用輾轉相除法就可以得到了
再來, 計算 n = pq
m, n 這兩個數便是 public key

編碼過程是, 若資料為 a, 將其看成是一個大整數, 假設 a < n
如果 a >= n 的話, 就將 a 表成 s 進位 (s <= n, 通常取 s = 2^t),
則每一位數均小於 n, 然後分段編碼
接下來, 計算 b == a^m mod n, (0 <= b < n),
b 就是編碼後的資料

解碼的過程是, 計算 c == b^r mod pq (0 <= c < pq),
於是乎, 解碼完畢 等會會證明 c 和 a 其實是相等的 :)

如果第三者進行竊聽時, 他會得到幾個數: m, n(=pq), b
他如果要解碼的話, 必須想辦法得到 r
所以, 他必須先對 n 作質因數分解
要防止他分解, 最有效的方法是找兩個非常的大質數 p, q,
使第三者作因數分解時發生困難
<定理>
若 p, q 是相異質數, rm == 1 mod (p-1)(q-1),
a 是任意一個正整數, b == a^m mod pq, c == b^r mod pq,
則 c == a mod pq

證明的過程, 會用到費馬小定理, 敘述如下:
m 是任一質數, n 是任一整數, 則 n^m == n mod m
(換另一句話說, 如果 n 和 m 互質, 則 n^(m-1) == 1 mod m)
運用一些基本的群論的知識, 就可以很容易地證出費馬小定理的

<證明>
因為 rm == 1 mod (p-1)(q-1), 所以 rm = k(p-1)(q-1) + 1, 其中 k 是整數
因為在 molo 中是 preserve 乘法的
(x == y mod z and u == v mod z => xu == yv mod z),
所以, c == b^r == (a^m)^r == a^(rm) == a^(k(p-1)(q-1)+1) mod pq

1. 如果 a 不是 p 的倍數, 也不是 q 的倍數時,
則 a^(p-1) == 1 mod p (費馬小定理) => a^(k(p-1)(q-1)) == 1 mod p
a^(q-1) == 1 mod q (費馬小定理) => a^(k(p-1)(q-1)) == 1 mod q
所以 p, q 均能整除 a^(k(p-1)(q-1)) - 1 => pq | a^(k(p-1)(q-1)) - 1
即 a^(k(p-1)(q-1)) == 1 mod pq
=> c == a^(k(p-1)(q-1)+1) == a mod pq

2. 如果 a 是 p 的倍數, 但不是 q 的倍數時,
則 a^(q-1) == 1 mod q (費馬小定理)
=> a^(k(p-1)(q-1)) == 1 mod q
=> c == a^(k(p-1)(q-1)+1) == a mod q
=> q | c - a
因 p | a
=> c == a^(k(p-1)(q-1)+1) == 0 mod p
=> p | c - a
所以, pq | c - a => c == a mod pq

3. 如果 a 是 q 的倍數, 但不是 p 的倍數時, 證明同上

4. 如果 a 同時是 p 和 q 的倍數時,
則 pq | a
=> c == a^(k(p-1)(q-1)+1) == 0 mod pq
=> pq | c - a
=> c == a mod pq
Q.E.D.

這個定理說明 a 經過編碼為 b 再經過解碼為 c 時, a == c mod n (n = pq)
但我們在做編碼解碼時, 限制 0 <= a < n, 0 <= c < n,
所以這就是說 a 等於 c, 所以這個過程確實能做到編碼解碼的功能

二、RSA 的安全性

RSA的安全性依賴於大數分解，但是否等同於大數分解一直未能得到理論上的證明，因為沒有證明破解
RSA就一定需要作大數分解。假設存在一種無須分解大數的演算法，那它肯定可以修改成為大數分解演算法。目前， RSA
的一些變種演算法已被證明等價於大數分解。不管怎樣，分解n是最顯然的攻擊方法。現在，人們已能分解多個十進制位的大素數。因此，模數n
必須選大一些，因具體適用情況而定。

三、RSA的速度

由於進行的都是大數計算，使得RSA最快的情況也比DES慢上倍，無論是軟體還是硬體實現。速度一直是RSA的缺陷。一般來說只用於少量數據加密。

四、RSA的選擇密文攻擊

RSA在選擇密文攻擊面前很脆弱。一般攻擊者是將某一信息作一下偽裝( Blind)，讓擁有私鑰的實體簽署。然後，經過計算就可得到它所想要的信息。實際上，攻擊利用的都是同一個弱點，即存在這樣一個事實：乘冪保留了輸入的乘法結構：

( XM )^d = X^d *M^d mod n

前面已經提到，這個固有的問題來自於公鑰密碼系統的最有用的特徵--每個人都能使用公鑰。但從演算法上無法解決這一問題，主要措施有兩條：一條是採用好的公
鑰協議，保證工作過程中實體不對其他實體任意產生的信息解密，不對自己一無所知的信息簽名；另一條是決不對陌生人送來的隨機文檔簽名，簽名時首先使用
One-Way HashFunction 對文檔作HASH處理，或同時使用不同的簽名演算法。在中提到了幾種不同類型的攻擊方法。

五、RSA的公共模數攻擊

若系統中共有一個模數，只是不同的人擁有不同的e和d，系統將是危險的。最普遍的情況是同一信息用不同的公鑰加密，這些公鑰共模而且互質，那末該信息無需私鑰就可得到恢復。設P為信息明文，兩個加密密鑰為e1和e2，公共模數是n，則：

C1 = P^e1 mod n

C2 = P^e2 mod n

密碼分析者知道n、e1、e2、C1和C2，就能得到P。

因為e1和e2互質，故用Euclidean演算法能找到r和s，滿足：

r * e1 + s * e2 = 1

假設r為負數，需再用Euclidean演算法計算C1^(-1)，則

( C1^(-1) )^(-r) * C2^s = P mod n

另外，還有其它幾種利用公共模數攻擊的方法。總之，如果知道給定模數的一對e和d，一是有利於攻擊者分解模數，一是有利於攻擊者計算出其它成對的e』和d』，而無需分解模數。解決辦法只有一個，那就是不要共享模數n。

RSA的小指數攻擊。 有一種提高 RSA速度的建議是使公鑰e取較小的值，這樣會使加密變得易於實現，速度有
所提高。但這樣作是不安全的，對付辦法就是e和d都取較大的值。

RSA演算法是
第一個能同時用於加密和數字簽名的演算法，也易於理解和操作。RSA是被研究得最廣泛的公鑰演算法，從提出到現在已近二十年，經歷了各種攻擊的考驗，逐漸為人
們接受，普遍認為是目前最優秀的公鑰方案之一。RSA的安全性依賴於大數的因子分解，但並沒有從理論上證明破譯RSA的難度與大數分解難度等價。即RSA
的重大缺陷是無法從理論上把握它的保密性能
如何，而且密碼學界多數人士傾向於因子分解不是NPC問題。
RSA的缺點主要有：A)產生密鑰很麻煩，受到素數產生技術的限制，因而難以做到一次一密。B)分組長度太大，為保證安全性，n 至少也要 600
bits
以上，使運算代價很高，尤其是速度較慢，較對稱密碼演算法慢幾個數量級；且隨著大數分解技術的發展，這個長度還在增加，不利於數據格式的標准化。目
前，SET( Secure Electronic Transaction )協議中要求CA採用比特長的密鑰，其他實體使用比特的密鑰。

C語言實現

#include <stdio.h>
int candp(int a,int b,int c)
{ int r=1;
b=b+1;
while(b!=1)
{
r=r*a;
r=r%c;
b--;
}
printf("%d\n",r);
return r;
}
void main()
{
int p,q,e,d,m,n,t,c,r;
char s;
printf("please input the p,q: ");
scanf("%d%d",&p,&q);
n=p*q;
printf("the n is %3d\n",n);
t=(p-1)*(q-1);
printf("the t is %3d\n",t);
printf("please input the e: ");
scanf("%d",&e);
if(e<1||e>t)
{
printf("e is error,please input again: ");
scanf("%d",&e);
}
d=1;
while(((e*d)%t)!=1) d++;
printf("then caculate out that the d is %d\n",d);
printf("the cipher please input 1\n");
printf("the plain please input 2\n");
scanf("%d",&r);
switch(r)
{
case 1: printf("input the m: "); /*輸入要加密的明文數字*/
scanf("%d",&m);
c=candp(m,e,n);
printf("the cipher is %d\n",c);break;
case 2: printf("input the c: "); /*輸入要解密的密文數字*/
scanf("%d",&c);
m=candp(c,d,n);
printf("the cipher is %d\n",m);break;
}
getch();
}

『叄』 用C語言來實現DES加密演算法（很急）兩天內

DES雖然不難但是挺繁復的，代碼如下，關鍵點都有英文解釋，仔細看。各個函數的功能都可以從函數名看出來。

#include "pch.h"
#include "misc.h"
#include "des.h"

NAMESPACE_BEGIN(CryptoPP)

/* Tables defined in the Data Encryption Standard documents
* Three of these tables, the initial permutation, the final
* permutation and the expansion operator, are regular enough that
* for speed, we hard-code them. They're here for reference only.
* Also, the S and P boxes are used by a separate program, gensp.c,
* to build the combined SP box, Spbox[]. They're also here just
* for reference.
*/
#ifdef notdef
/* initial permutation IP */
static byte ip[] = {
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
};

/* final permutation IP^-1 */
static byte fp[] = {
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
};
/* expansion operation matrix */
static byte ei[] = {
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
};
/* The (in)famous S-boxes */
static byte sbox[8][64] = {
/* S1 */
14, 4, 13, 1, 2, 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,

/* S2 */
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,

/* S3 */
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,

/* S4 */
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,

/* S5 */
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,

/* S6 */
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,

/* S7 */
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,

/* S8 */
13, 2, 8, 4, 6, 15, 11, 1, 10, 9, 3, 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
};

/* 32-bit permutation function P used on the output of the S-boxes */
static byte p32i[] = {
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
};
#endif

/* permuted choice table (key) */
static const byte pc1[] = {
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
};

/* number left rotations of pc1 */
static const byte totrot[] = {
1,2,4,6,8,10,12,14,15,17,19,21,23,25,27,28
};

/* permuted choice key (table) */
static const byte pc2[] = {
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
};

/* End of DES-defined tables */

/* bit 0 is left-most in byte */
static const int bytebit[] = {
0200,0100,040,020,010,04,02,01
};

/* Set key (initialize key schele array) */
DES::DES(const byte *key, CipherDir dir)
: k(32)
{
SecByteBlock buffer(56+56+8);
byte *const pc1m=buffer; /* place to modify pc1 into */
byte *const pcr=pc1m+56; /* place to rotate pc1 into */
byte *const ks=pcr+56;
register int i,j,l;
int m;

for (j=0; j<56; j++) { /* convert pc1 to bits of key */
l=pc1[j]-1; /* integer bit location */
m = l & 07; /* find bit */
pc1m[j]=(key[l>>3] & /* find which key byte l is in */
bytebit[m]) /* and which bit of that byte */
? 1 : 0; /* and store 1-bit result */
}
for (i=0; i<16; i++) { /* key chunk for each iteration */
memset(ks,0,8); /* Clear key schele */
for (j=0; j<56; j++) /* rotate pc1 the right amount */
pcr[j] = pc1m[(l=j+totrot[i])<(j<28? 28 : 56) ? l: l-28];
/* rotate left and right halves independently */
for (j=0; j<48; j++){ /* select bits indivially */
/* check bit that goes to ks[j] */
if (pcr[pc2[j]-1]){
/* mask it in if it's there */
l= j % 6;
ks[j/6] |= bytebit[l] >> 2;
}
}
/* Now convert to odd/even interleaved form for use in F */
k[2*i] = ((word32)ks[0] << 24)
| ((word32)ks[2] << 16)
| ((word32)ks[4] << 8)
| ((word32)ks[6]);
k[2*i+1] = ((word32)ks[1] << 24)
| ((word32)ks[3] << 16)
| ((word32)ks[5] << 8)
| ((word32)ks[7]);
}

if (dir==DECRYPTION) // reverse key schele order
for (i=0; i<16; i+=2)
{
std::swap(k[i], k[32-2-i]);
std::swap(k[i+1], k[32-1-i]);
}
}
/* End of C code common to both versions */

/* C code only in portable version */

// Richard Outerbridge's initial permutation algorithm
/*
inline void IPERM(word32 &left, word32 &right)
{
word32 work;

work = ((left >> 4) ^ right) & 0x0f0f0f0f;
right ^= work;
left ^= work << 4;
work = ((left >> 16) ^ right) & 0xffff;
right ^= work;
left ^= work << 16;
work = ((right >> 2) ^ left) & 0x33333333;
left ^= work;
right ^= (work << 2);
work = ((right >> 8) ^ left) & 0xff00ff;
left ^= work;
right ^= (work << 8);
right = rotl(right, 1);
work = (left ^ right) & 0xaaaaaaaa;
left ^= work;
right ^= work;
left = rotl(left, 1);
}
inline void FPERM(word32 &left, word32 &right)
{
word32 work;

right = rotr(right, 1);
work = (left ^ right) & 0xaaaaaaaa;
left ^= work;
right ^= work;
left = rotr(left, 1);
work = ((left >> 8) ^ right) & 0xff00ff;
right ^= work;
left ^= work << 8;
work = ((left >> 2) ^ right) & 0x33333333;
right ^= work;
left ^= work << 2;
work = ((right >> 16) ^ left) & 0xffff;
left ^= work;
right ^= work << 16;
work = ((right >> 4) ^ left) & 0x0f0f0f0f;
left ^= work;
right ^= work << 4;
}
*/

// Wei Dai's modification to Richard Outerbridge's initial permutation
// algorithm, this one is faster if you have access to rotate instructions
// (like in MSVC)
inline void IPERM(word32 &left, word32 &right)
{
word32 work;

right = rotl(right, 4U);
work = (left ^ right) & 0xf0f0f0f0;
left ^= work;
right = rotr(right^work, 20U);
work = (left ^ right) & 0xffff0000;
left ^= work;
right = rotr(right^work, 18U);
work = (left ^ right) & 0x33333333;
left ^= work;
right = rotr(right^work, 6U);
work = (left ^ right) & 0x00ff00ff;
left ^= work;
right = rotl(right^work, 9U);
work = (left ^ right) & 0xaaaaaaaa;
left = rotl(left^work, 1U);
right ^= work;
}

inline void FPERM(word32 &left, word32 &right)
{
word32 work;

right = rotr(right, 1U);
work = (left ^ right) & 0xaaaaaaaa;
right ^= work;
left = rotr(left^work, 9U);
work = (left ^ right) & 0x00ff00ff;
right ^= work;
left = rotl(left^work, 6U);
work = (left ^ right) & 0x33333333;
right ^= work;
left = rotl(left^work, 18U);
work = (left ^ right) & 0xffff0000;
right ^= work;
left = rotl(left^work, 20U);
work = (left ^ right) & 0xf0f0f0f0;
right ^= work;
left = rotr(left^work, 4U);
}

// Encrypt or decrypt a block of data in ECB mode
void DES::ProcessBlock(const byte *inBlock, byte * outBlock) const
{
word32 l,r,work;

#ifdef IS_LITTLE_ENDIAN
l = byteReverse(*(word32 *)inBlock);
r = byteReverse(*(word32 *)(inBlock+4));
#else
l = *(word32 *)inBlock;
r = *(word32 *)(inBlock+4);
#endif

IPERM(l,r);

const word32 *kptr=k;

for (unsigned i=0; i<8; i++)
{
work = rotr(r, 4U) ^ kptr[4*i+0];
l ^= Spbox[6][(work) & 0x3f]
^ Spbox[4][(work >> 8) & 0x3f]
^ Spbox[2][(work >> 16) & 0x3f]
^ Spbox[0][(work >> 24) & 0x3f];
work = r ^ kptr[4*i+1];
l ^= Spbox[7][(work) & 0x3f]
^ Spbox[5][(work >> 8) & 0x3f]
^ Spbox[3][(work >> 16) & 0x3f]
^ Spbox[1][(work >> 24) & 0x3f];

work = rotr(l, 4U) ^ kptr[4*i+2];
r ^= Spbox[6][(work) & 0x3f]
^ Spbox[4][(work >> 8) & 0x3f]
^ Spbox[2][(work >> 16) & 0x3f]
^ Spbox[0][(work >> 24) & 0x3f];
work = l ^ kptr[4*i+3];
r ^= Spbox[7][(work) & 0x3f]
^ Spbox[5][(work >> 8) & 0x3f]
^ Spbox[3][(work >> 16) & 0x3f]
^ Spbox[1][(work >> 24) & 0x3f];
}

FPERM(l,r);

#ifdef IS_LITTLE_ENDIAN
*(word32 *)outBlock = byteReverse(r);
*(word32 *)(outBlock+4) = byteReverse(l);
#else
*(word32 *)outBlock = r;
*(word32 *)(outBlock+4) = l;
#endif
}

void DES_EDE_Encryption::ProcessBlock(byte *inoutBlock) const
{
e.ProcessBlock(inoutBlock);
d.ProcessBlock(inoutBlock);
e.ProcessBlock(inoutBlock);
}

void DES_EDE_Encryption::ProcessBlock(const byte *inBlock, byte *outBlock) const
{
e.ProcessBlock(inBlock, outBlock);
d.ProcessBlock(outBlock);
e.ProcessBlock(outBlock);
}

void DES_EDE_Decryption::ProcessBlock(byte *inoutBlock) const
{
d.ProcessBlock(inoutBlock);
e.ProcessBlock(inoutBlock);
d.ProcessBlock(inoutBlock);
}

void DES_EDE_Decryption::ProcessBlock(const byte *inBlock, byte *outBlock) const
{
d.ProcessBlock(inBlock, outBlock);
e.ProcessBlock(outBlock);
d.ProcessBlock(outBlock);
}

void TripleDES_Encryption::ProcessBlock(byte *inoutBlock) const
{
e1.ProcessBlock(inoutBlock);
d.ProcessBlock(inoutBlock);
e2.ProcessBlock(inoutBlock);
}

void TripleDES_Encryption::ProcessBlock(const byte *inBlock, byte *outBlock) const
{
e1.ProcessBlock(inBlock, outBlock);
d.ProcessBlock(outBlock);
e2.ProcessBlock(outBlock);
}

void TripleDES_Decryption::ProcessBlock(byte *inoutBlock) const
{
d1.ProcessBlock(inoutBlock);
e.ProcessBlock(inoutBlock);
d2.ProcessBlock(inoutBlock);
}

void TripleDES_Decryption::ProcessBlock(const byte *inBlock, byte *outBlock) const
{
d1.ProcessBlock(inBlock, outBlock);
e.ProcessBlock(outBlock);
d2.ProcessBlock(outBlock);
}

『肆』 用C語言編程實現CANNY演算法

創建一個名字為canny.par的文件，就同你建立一個test.txt一樣