- 密钥长度 128 位
- 分组长度 64位 (8字节的倍数)
- TEA系列算法典型特征是采用密钥调度常数
0x9e3779b9

#include <stdio.h>
void encrypt(unsigned int* v, unsigned int* key) {
unsigned int l = v[0], r = v[1], sum = 0, delta = 0x9e3779b9;
for (size_t i = 0; i < 32; i++) {
sum += delta;
l += ((r << 4) + key[0]) ^ (r + sum) ^ ((r >> 5) + key[1]);
r += ((l << 4) + key[2]) ^ (l + sum) ^ ((l >> 5) + key[3]);
}
v[0] = l;
v[1] = r;
}
void decrypt(unsigned int* v, unsigned int* key) {
unsigned int l = v[0], r = v[1], sum = 0, delta = 0x9e3779b9;
sum = delta *32;
for (size_t i = 0; i < 32; i++) {
r -= ((l << 4) + key[2]) ^ (l + sum) ^ ((l >> 5) + key[3]);
l -= ((r << 4) + key[0]) ^ (r + sum) ^ ((r >> 5) + key[1]);
sum -= delta;
}
v[0] = l;
v[1] = r;
}
int main(int argc, char const *argv[])
{
//test
unsigned int v[2]={1,2},key[4]={1,2,3,4};
printf("%u,%u\\n",v[0],v[1]);
encrypt(v,key);
printf("%u,%u\\n",v[0],v[1]);
decrypt(v,key);
printf("%u,%u\\n",v[0],v[1]);
return 0;
}

#include <stdio.h>
void encrypt(unsigned int* v, unsigned int* key) {
unsigned int l = v[0], r = v[1], sum = 0, delta = 0x9e3779b9;
for (size_t i = 0; i < 32; i++) {
l += (((r << 4) ^ (r >> 5)) + r) ^ (sum + key[sum & 3]);
sum += delta;
r += (((l << 4) ^ (l >> 5)) + l) ^ (sum + key[(sum >> 11) & 3]);
}
v[0] = l;
v[1] = r;
}
void decrypt(unsigned int* v, unsigned int* key) {
unsigned int l = v[0], r = v[1], sum = 0, delta = 0x9e3779b9;
sum = delta * 32;
for (size_t i = 0; i < 32; i++) {
r -= (((l << 4) ^ (l >> 5)) + l) ^ (sum + key[(sum >> 11) & 3]);
sum -= delta;
l -= (((r << 4) ^ (r >> 5)) + r) ^ (sum + key[sum & 3]);
}
v[0] = l;
v[1] = r;
}
int main(int argc, char const *argv[])
{
//test
unsigned int v[2]={1,2},key[4]={1,2,3,4};
printf("%u,%u\\n",v[0],v[1]);
encrypt(v,key);
printf("%u,%u\\n",v[0],v[1]);
decrypt(v,key);
printf("%u,%u\\n",v[0],v[1]);
return 0;
}

#include <stdbool.h>
#include <stdio.h>
#define MX \\
((z >> 5 ^ y << 2) + (y >> 3 ^ z << 4) ^ (sum ^ y) + (k[p & 3 ^ e] ^ z))
bool btea(unsigned int* v, int n, unsigned int* k) {
unsigned int z = v[n - 1], y = v[0], sum = 0, e, DELTA = 0x9e3779b9;
unsigned int p, q;
if (n > 1) { /* Coding Part */
q = 6 + 52 / n;
while (q-- > 0) {
sum += DELTA;
e = (sum >> 2) & 3;
for (p = 0; p < n - 1; p++)
y = v[p + 1], z = v[p] += MX;
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;
}
int main(int argc, char const* argv[]) {
// test
unsigned int v[2] = {1, 2}, key[4] = {1, 2, 3, 4};
printf("%u,%u\\n", v[0], v[1]);
btea(v, 2, key);
printf("%u,%u\\n", v[0], v[1]);
btea(v, -2, key);
printf("%u,%u\\n", v[0], v[1]);
return 0;
}