XXTEA加密算法的KEIL C實現(xiàn)

?

#include ?
#include ?
#include ?
#include ?

#define MX (z>>5^y<<2) + (y>>3^z<<4)^(sum^y) + (k[p&3^e]^z)

#define??delta?????0x9e3779b9

/* 注意：delta的取值是隨機的，但是為了避免不良的取值，采取的是
黃金分割數(shù)（根號5－2）/2與2的32次方的乘積。為0x9e3779b9。???
在解密中，sum＝delta×round，如：delta×32＝13C6EF3720*/?????
//unsigned?long?buffer[];
//unsigned?long?keybuffer[];?

long?btea(long*?v,char?n,long*?k);

long ?btea(long* v, char?n, long* k)?
{?
??? unsigned long z=v[n-1], y=v[0], sum=0, e, DELTA=0x9e3779b9;
??? long p, q ;
??? if (n > 1)?
?{??????????
?/* Coding Part */
????? q = 6 + 52/n;
????? while (q-- > 0) {
??????? sum += DELTA;
??????? e = (sum >> 2) & 3;
??????? for (p=0; p ??????? y = v[0];
??????? z = v[n-1] += MX;
????? }
????? return 0 ;?
??? }?
?else if (n < -1)?
?{??
? /* Decoding Part */
????? n = -n;
????? q = 6 + 52/n;
????? sum = q*DELTA ;
????? while (sum != 0) {
??????? e = (sum >> 2) & 3;
??????? for (p=n-1; p>0; p--) z = v[p-1], y = v[p] -= MX;
??????? z = v[n-1];
??????? y = v[0] -= MX;
??????? sum -= DELTA;
????? }
????? return 0;
??? }
??? return 1;
}

?void ?main()
?{
?unsigned?long?plainbuffer[2];
?unsigned?long?testbuffer[4];
?unsigned?long?plain[2]={0x23FF28AA,0xA76B4B04 };
?unsigned?long?test[4]={0x01122334,0x45566778,0x899AABBC,0xCDDEEFF0};
?unsigned?long?key[4]={0x01234567,0x89ABCDEF,0x01234567,0x89ABCDEF};

?while(1)
?{
??memcpy(plainbuffer,plain,8);
??memcpy(testbuffer,test,16);
??btea(plainbuffer, 2,key);
??btea(plainbuffer, -2,key);
??btea(testbuffer, 4,key);
??btea(testbuffer, -4,key);
??_nop_;
?//?btea(buffer, -2,keybuffer);

??}
}