Javascript 不同语言中相同代码的不同行为
我正在尝试用Javascript(Node.js)解密。我尝试在C++中做同样的事情,它的工作与预期一样:Javascript 不同语言中相同代码的不同行为,javascript,c++,xxtea,Javascript,C++,Xxtea,我正在尝试用Javascript(Node.js)解密。我尝试在C++中做同样的事情,它的工作与预期一样: #define DELTA 0x9e3779b9 #define MX (((z>>5^y<<2) + (y>>3^z<<4)) ^ ((sum^y) + (key[(p&3)^e] ^ z))) #define CRYPT_WORDS (64-4)/4 #define CRYPT_OFFSET 1 void btea_decry
#define DELTA 0x9e3779b9
#define MX (((z>>5^y<<2) + (y>>3^z<<4)) ^ ((sum^y) + (key[(p&3)^e] ^ z)))
#define CRYPT_WORDS (64-4)/4
#define CRYPT_OFFSET 1
void btea_decrypt(uint32_t *v, int n, int base_rounds, uint32_t const key[4])
{
uint32_t y, z, sum;
unsigned p, rounds, e;
/* Decoding Part */
rounds = base_rounds + 52/n;
sum = rounds*DELTA;
y = v[0];
do {
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;
} while (--rounds);
}
int main()
{
static const uint32_t key[4] = {0x875bcc51, 0xa7637a66, 0x50960967, 0xf8536c51};
uint32_t buf[64] = {16, 23, 163, 242, 214, 213, 125, 48, 167, 44, 232,
23, 160, 192, 244, 116, 38, 255, 200, 38, 43, 57,
18, 235, 206, 103, 161, 210, 187, 164, 42, 227, 139,
248, 141, 205, 51, 132, 115, 233, 39, 53, 136, 207,
238, 190, 111, 57, 117, 233, 67, 133, 165, 84, 154,
161, 165, 173, 76, 115, 108, 0, 0, 71};
uint32_t cryptpart[CRYPT_WORDS];
// Decrypt encrypted portion
for (int i = 0; i < CRYPT_WORDS; i++) {
cryptpart[i] =
((uint32_t)buf[CRYPT_OFFSET+4*i ]) << 0 |
((uint32_t)buf[CRYPT_OFFSET+4*i+1]) << 8 |
((uint32_t)buf[CRYPT_OFFSET+4*i+2]) << 16 |
((uint32_t)buf[CRYPT_OFFSET+4*i+3]) << 24;
}
btea_decrypt(cryptpart, CRYPT_WORDS, 1, key);
for (int i = 0; i < CRYPT_WORDS; i++) {
buf[CRYPT_OFFSET+4*i ] = cryptpart[i] >> 0;
buf[CRYPT_OFFSET+4*i+1] = cryptpart[i] >> 8;
buf[CRYPT_OFFSET+4*i+2] = cryptpart[i] >> 16;
buf[CRYPT_OFFSET+4*i+3] = cryptpart[i] >> 24;
}
for (const auto& e : buf) {
std::cout << e << ", ";
}
}
但是在Node.js中使用相同的代码(ported):
function btea_decrypt(v, n, base_rounds, key)
{
let y, z, sum;
let p, rounds, e;
/* Decoding Part */
rounds = base_rounds + 52/n;
sum = rounds*0x9e3779b9;
y = v[0];
do {
e = (sum >> 2) & 3;
for (p=n-1; p>0; p--) {
z = v[p-1];
y = v[p] -= (((z>>5^y<<2) + (y>>3^z<<4)) ^ ((sum^y) + (key[(p&3)^e] ^ z)));
}
z = v[n-1];
y = v[0] -= (((z>>5^y<<2) + (y>>3^z<<4)) ^ ((sum^y) + (key[(p&3)^e] ^ z)));
sum -= 0x9e3779b9;
} while (--rounds);
return v;
}
function main() {
let key = [0x875bcc51, 0xa7637a66, 0x50960967, 0xf8536c51];
let buf = [16, 23, 163, 242, 214, 213, 125, 48, 167, 44, 232,
23, 160, 192, 244, 116, 38, 255, 200, 38, 43, 57,
18, 235, 206, 103, 161, 210, 187, 164, 42, 227, 139,
248, 141, 205, 51, 132, 115, 233, 39, 53, 136, 207,
238, 190, 111, 57, 117, 233, 67, 133, 165, 84, 154,
161, 165, 173, 76, 115, 108, 0, 0, 71];
let cryptpart = [];
// Decrypt encrypted portion
for (let i = 0; i < (64-4)/4; i++) {
cryptpart[i] =
(buf[1+4*i ]) << 0 |
(buf[1+4*i+1]) << 8 |
(buf[1+4*i+2]) << 16 |
(buf[1+4*i+3]) << 24;
}
cryptpart = btea_decrypt(cryptpart, (64-4)/4, 1, key);
for (let i = 0; i < (64-4)/4; i++) {
buf[1+4*i ] = cryptpart[i] >> 0;
buf[1+4*i+1] = cryptpart[i] >> 8;
buf[1+4*i+2] = cryptpart[i] >> 16;
buf[1+4*i+3] = cryptpart[i] >> 24;
}
console.log(buf)
}
函数btea_decrypt(v,n,base_轮,key)
{
让y,z,和;
设p,轮数,e;
/*解码部分*/
轮数=基本轮数+52/n;
总和=轮数*0x9e3779b9;
y=v[0];
做{
e=(总和>>2)和3;
对于(p=n-1;p>0;p--){
z=v[p-1];
y=v[p]-=((z>>5^y3^z5^y3^z16;
buf[1+4*i+3]=cryptpart[i]>>24;
}
控制台日志(buf)
}
它仍然卡在do…中,而将永远循环
我想,JavaScript和C++处理代码< 0x9E379B9</C> >不同,因为C++中的代码> 0x9E379B9* 15 < /Cord>在C++中等于39816536535和1161830871。C++中的数学错误,以及如何在JS?
中实现这一点。
如果我的英语不是最好的,很抱歉。您的问题是由整数溢出引起的。unint32是一个固定大小为2^32位的整数。0x9e3779b9*15是39816536535,约为2^35
这意味着您会出现溢出,因为内存位置根本不足以容纳您的数字。Javascript没有这个问题,因为它不是静态类型的,并且内存中分配的大小将动态增加以容纳它
< >使用C++的较大数据类型,如<代码>未签名的长< /COD>或<代码> SiZeZt<<代码>(大多数系统中的<代码>未签名的长< /COD>),最好使用<代码> Auto <代码>,让编译器为您决定:
auto sum=rounds*0x9e3779b9;
这将解决您的问题,并确保总和足以容纳数字
<>边注释:C++中有很多C,我们在C++中尝试避免< <代码> >定义< /C> >变量,而不是在变量的顶部。当C++代码更喜欢<代码> const AutoX[/Cuff] >而不是<代码> >定义x<代码>(可能是代码> conxPRP< /COD>如果是基本类型)C++有64位整数类型;JavaScript没有。现代JavaScript支持“大整数”键入。您正在经历的被称为整数溢出。不确定JS是否会这样做。在JS中,由于赋值中的除法,rounds
被解释为浮点,因此可能永远不会为0-因此循环永远不会结束。请使用Math.floor()
,Math.ceil()
,Math.round()
,或Math.trunc()
从四舍五入中获取整数。
function btea_decrypt(v, n, base_rounds, key)
{
let y, z, sum;
let p, rounds, e;
/* Decoding Part */
rounds = base_rounds + 52/n;
sum = rounds*0x9e3779b9;
y = v[0];
do {
e = (sum >> 2) & 3;
for (p=n-1; p>0; p--) {
z = v[p-1];
y = v[p] -= (((z>>5^y<<2) + (y>>3^z<<4)) ^ ((sum^y) + (key[(p&3)^e] ^ z)));
}
z = v[n-1];
y = v[0] -= (((z>>5^y<<2) + (y>>3^z<<4)) ^ ((sum^y) + (key[(p&3)^e] ^ z)));
sum -= 0x9e3779b9;
} while (--rounds);
return v;
}
function main() {
let key = [0x875bcc51, 0xa7637a66, 0x50960967, 0xf8536c51];
let buf = [16, 23, 163, 242, 214, 213, 125, 48, 167, 44, 232,
23, 160, 192, 244, 116, 38, 255, 200, 38, 43, 57,
18, 235, 206, 103, 161, 210, 187, 164, 42, 227, 139,
248, 141, 205, 51, 132, 115, 233, 39, 53, 136, 207,
238, 190, 111, 57, 117, 233, 67, 133, 165, 84, 154,
161, 165, 173, 76, 115, 108, 0, 0, 71];
let cryptpart = [];
// Decrypt encrypted portion
for (let i = 0; i < (64-4)/4; i++) {
cryptpart[i] =
(buf[1+4*i ]) << 0 |
(buf[1+4*i+1]) << 8 |
(buf[1+4*i+2]) << 16 |
(buf[1+4*i+3]) << 24;
}
cryptpart = btea_decrypt(cryptpart, (64-4)/4, 1, key);
for (let i = 0; i < (64-4)/4; i++) {
buf[1+4*i ] = cryptpart[i] >> 0;
buf[1+4*i+1] = cryptpart[i] >> 8;
buf[1+4*i+2] = cryptpart[i] >> 16;
buf[1+4*i+3] = cryptpart[i] >> 24;
}
console.log(buf)
}