算法簡單,而且效率高,每次可以操作8個字節(jié)的數(shù)據(jù),加密解密的KEY為16字節(jié),即包含4個int數(shù)據(jù)的int型數(shù)組,加密輪數(shù)應(yīng)為8的倍數(shù),一般比較常用的輪數(shù)為64,32,16,QQ原來就是用TEA16來還原密碼的.

TEA算法

核心為：

#include

void encrypt (uint32_t* v, uint32_t* k) {

uint32_t v0=v[0], v1=v[1], sum=0, i; /* set up */

uint32_t delta=0x9e3779b9; /* a key schedule constant */

uint32_t k0=k[0], k1=k[1], k2=k[2], k3=k[3]; /* cache key */

for (i=0; i < 32; i++) { /* basic cycle start */

sum += delta;

v0 += ((v1<<4) + k0) ^ (v1 + sum) ^ ((v1>>5) + k1);

v1 += ((v0<<4) + k2) ^ (v0 + sum) ^ ((v0>>5) + k3);

} /* end cycle */

v[0]=v0; v[1]=v1;

}

void decrypt (uint32_t* v, uint32_t* k) {

uint32_t v0=v[0], v1=v[1], sum=0xC6EF3720, i; /* set up */

uint32_t delta=0x9e3779b9; /* a key schedule constant */

uint32_t k0=k[0], k1=k[1], k2=k[2], k3=k[3]; /* cache key */

for (i=0; i<32; i++) { /* basic cycle start */

v1 -= ((v0<<4) + k2) ^ (v0 + sum) ^ ((v0>>5) + k3);

v0 -= ((v1<<4) + k0) ^ (v1 + sum) ^ ((v1>>5) + k1);

sum -= delta;

} /* end cycle */

v[0]=v0; v[1]=v1;

}

PHP部分代碼非我原創(chuàng)，大家可以了解一下這方面的知識

上面的是TEA的算法，XTEA的算法為：

#include

void encipher(unsigned int num_rounds, uint32_t v[2], uint32_t const k[4]) {

unsigned int i;

uint32_t v0=v[0], v1=v[1], sum=0, delta=0x9E3779B9;

for (i=0; i < num_rounds; i++) {

v0 += (((v1 << 4) ^ (v1 >> 5)) + v1) ^ (sum + k[sum & 3]);

sum += delta;

v1 += (((v0 << 4) ^ (v0 >> 5)) + v0) ^ (sum + k[(sum>>11) & 3]);

}

v[0]=v0; v[1]=v1;

}

void decipher(unsigned int num_rounds, uint32_t v[2], uint32_t const k[4]) {

unsigned int i;

uint32_t v0=v[0], v1=v[1], delta=0x9E3779B9, sum=delta*num_rounds;

for (i=0; i < num_rounds; i++) {

v1 ?= (((v0 << 4) ^ (v0 >> 5)) + v0) ^ (sum + k[(sum>>11) & 3]);

sum ?= delta;

v0 ?= (((v1 << 4) ^ (v1 >> 5)) + v1) ^ (sum + k[sum & 3]);

}

v[0]=v0; v[1]=v1;

}

那PHP中只需要把運算的位置改下就OK

private function _teaencipherLong($y, $z, &$w, &$k) {

$sum = ( integer ) 0;

$delta = 0x9E3779B9;

$n = ( integer ) $this->n_iter;

while ( $n -- > 0 ) {

$y = $this->_add ( $y, $this->_add ( $z << 4 ^ $this->_rshift ( $z, 5 ), $z ) ^ $this->_add ( $sum, $k [$sum & 3] ) );

$sum = $this->_add ( $sum, $delta );

$z = $this->_add ( $z, $this->_add ( $y << 4 ^ $this->_rshift ( $y, 5 ), $y ) ^ $this->_add ( $sum, $k [$this->_rshift ( $sum, 11 ) & 3] ) );

}

$w [0] = $y;

$w [1] = $z;

}

private function _decipherLong($y, $z, &$w, &$k) {

// sum = delta<<5, in general sum = delta * n

$sum = 0xC6EF3720;

$delta = 0x9E3779B9;

$n = ( integer ) $this->n_iter;

while ( $n -- > 0 ) {

$z = $this->_add ( $z, - ($this->_add ( $y << 4 ^ $this->_rshift ( $y, 5 ), $y ) ^ $this->_add ( $sum, $k [$this->_rshift ( $sum, 11 ) & 3] )) );

$sum = $this->_add ( $sum, - $delta );

$y = $this->_add ( $y, - ($this->_add ( $z << 4 ^ $this->_rshift ( $z, 5 ), $z ) ^ $this->_add ( $sum, $k [$sum & 3] )) );

}

$w [0] = $y;

$w [1] = $z;

}

XXTEA的算法

核心為

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

long btea(long* v, long 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;

}

也是運算不一樣，這個就不寫了，有人已經(jīng)寫過這方面的代碼了

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