我在将一些C代码翻译成Python时遇到了麻烦。我已尽力模拟C中整数溢出的性质,但到目前为止还没有成功。如果有人有一些见解,那将是非常感谢,因为我是python的新手。加密&解密如下:
iterations = 32
delta = 0x9e3779b9
xTeaKey = [<some int32>, <some int32>, <some int32>, <some int32>]
def int_overflow(val):
maxint = 2147483647
if not -maxint-1 <= val <= maxint:
val = (val + (maxint + 1)) % (2 * (maxint + 1)) - maxint - 1
if not -maxint-1 <= val <= maxint:
print "AAAAAH"
return val
def xTeaShuffle(x, sum, sumOffset) :
e1 = (x << 4) & 0xffffffff
e2 = x >> 5
e3 = e1 ^ e2
e4 = int_overflow(e3 + x)
e5a = int_overflow(sum + xTeaKey[(sum & 0x03)]);
e5b = int_overflow(sum + xTeaKey[((sum >> 11) & 0x03)]);
e5 = e5b if sumOffset else e5a
result = e4 ^ e5
return result
def xTeaEncode(data, length) :
i = 0
while i < length:
sum = 0
x1 = (data[i] << 16) + data[i + 1]
x2 = (data[i + 2] << 16) + data[i + 3]
iter = iterations
while iter > 0 :
x1 = int_overflow(x1 + xTeaShuffle(x2, sum, False))
sum = int_overflow(sum + delta);
x2 = int_overflow(x2 + xTeaShuffle(x1, sum, True))
iter -= 1
data[i] = (x1 >> 16) & 0xffff
data[i + 1] = x1 & 0xffff
data[i + 2] = (x2 >> 16) & 0xffff
data[i + 3] = x2 & 0xffff
i += 4
return
def xTeaDecode(data, length) :
i = 0
while i < length:
sum = int_overflow(delta * iterations)
x1 = (data[i] << 16) + data[i + 1]
x2 = (data[i + 2] << 16) + data[i + 3]
while (sum != 0) :
x2 = int_overflow(x2 - xTeaShuffle(x1, sum, True))
sum = int_overflow(sum - delta)
x1 = int_overflow(x1 - xTeaShuffle(x2, sum, False))
data[i] = (x1 >> 16) & 0xffff
data[i + 1] = x1 & 0xffff
data[i + 2] = (x2 >> 16) & 0xffff
data[i + 3] = x2 & 0xffff
i += 4
return
原始C代码
DllExport void _stdcall XTEAEncode16(unsigned short *data, unsigned char dataLength)
{
unsigned char i = 0;
int x1;
int x2;
int sum;
unsigned char iterationCount;
while (i < dataLength)
{
sum = 0;
x1 = ((unsigned int)data[i] << 16) + (unsigned int)data[i+1];
x2 = ((unsigned int)data[i+2] << 16) + (unsigned int)data[i+3];
iterationCount = NUM_ITERATIONS;
while (iterationCount > 0)
{
x1 += (((x2 << 4) ^ (x2 >> 5)) + x2) ^ (sum + XTEAKey[(sum & 0x03)]);
sum += DELTA;
x2 += (((x1 << 4) ^ (x1 >> 5)) + x1) ^ (sum + XTEAKey[((sum >> 11) & 0x03)]);
iterationCount--;
}
data[i] = (unsigned short)((unsigned int)x1>>16); /* take upper half as an int*/
data[i+1] = (unsigned short)(unsigned int)x1; /* take lower half */
data[i+2] = (unsigned short)((unsigned int)x2>>16); /* take upper half as an int*/
data[i+3] = (unsigned short)(unsigned int)x2; /* take lower half */
i += 4;
}
}
/**
* Decodes (deciphers) data.
* Note that data length must be a multiple of 4 words (64 bit).
*//* *< 16-bit data array *//* *< length of data array */
DllExport void _stdcall XTEADecode16(unsigned short* data, unsigned char dataLength )
{
unsigned char i = 0;
int x1;
int x2;
int sum;
unsigned char iterations;
iterations = NUM_ITERATIONS;
while (i < dataLength)
{
sum = DELTA * iterations;
x1 = ((unsigned int)data[i] << 16) + (unsigned int)data[i+1];
x2 = ((unsigned int)data[i+2] << 16) + (unsigned int)data[i+3];
while (sum != 0)
{
x2 -= (((x1 << 4) ^ (x1 >> 5)) + x1) ^ (sum + XTEAKey[((sum >> 11) & 0x03)]);
sum -= DELTA;
x1 -= (((x2 << 4) ^ (x2 >> 5)) + x2) ^ (sum + XTEAKey[(sum & 0x03)]);
}
data[i] = (unsigned short)((unsigned int)x1 >> 16); /* take upper half as an int*/
data[i+1] = (unsigned short)((unsigned int)x1); /* take lower half */
data[i+2] = (unsigned short)((unsigned int)x2 >> 16); /* take upper half as an int*/
data[i+3] = (unsigned short)((unsigned int)x2); /* take lower half */
i += 4;
}
}
从我的实验中,加密似乎没问题,但我永远无法解密到正确的值。任何见解将不胜感激。
答案 0 :(得分:0)
解决了!!!这是问题所在:
当向右移位(并且取决于实现)时,位可以用0或1填充,具体取决于最左边的位。这是为了保留两个恭维二进制数的符号。因此,当模拟c整数上的移位时,如果我们不模拟符号,我们必须用左边填充1,如果这是最左边的位。