C++ 莫顿码的计算
我正在尝试(为了计算莫顿码)将2个有符号的长数字(如C++ 莫顿码的计算,c++,math,bit-manipulation,64-bit,z-order-curve,C++,Math,Bit Manipulation,64 Bit,Z Order Curve,我正在尝试(为了计算莫顿码)将2个有符号的长数字(如x和y(32位))与值交错 案例1: x = 10; //1010 y = 10; //1010 结果将是: 11001100 案例2: x = -10; y = 10; 二进制表示是 x = 1111111111111111111111111111111111111111111111111111111111110110 y = 1010 对于交织,我只考虑32位表示,其中我可以将x的31位与y的31位交织, 使用以下代
x
和y
(32位))与值交错
案例1:
x = 10; //1010
y = 10; //1010
结果将是:
11001100
案例2:
x = -10;
y = 10;
二进制表示是
x = 1111111111111111111111111111111111111111111111111111111111110110
y = 1010
对于交织,我只考虑32位表示,其中我可以将x
的31位与y
的31位交织,
使用以下代码
signed long long x_y;
for (int i = 31; i >= 0; i--)
{
unsigned long long xbit = ((unsigned long) x)& (1 << i);
x_y|= (xbit << i);
unsigned long long ybit = ((unsigned long) y)& (1 << i);
if (i != 0)
{
x_y|= (x_y<< (i - 1));
}
else
{
(x_y= x_y<< 1) |= ybit;
}
}
有符号长x_y;
对于(int i=31;i>=0;i--)
{
unsigned long xbit=((unsigned long)x)和(1我认为下面的代码符合您的要求
莫顿码是64位的,我们用两个32位的数字交错生成64位的数字。
由于数字已经签名,我们必须考虑负数为,
if (x < 0) //value will be represented as 2's compliment,hence uses all 64 bits
{
value = x; //value is of 32 bit,so use only first lower 32 bits
cout << value;
value &= ~(1 << 31); //make sign bit to 0,as it does not contribute to real value.
}
我希望这有帮助。:)
unsigned long long x_y_copy = 0; //make a copy of ur morton code
//looping for each bit of two 32 bit numbers starting from MSB.
for (int i = 31; i >=0; i--)
{
//making mort to 0,so because shifting causes loss of data
mort = 0;
//take 32 bit from x
int xbit = ((unsigned long)x)& (1 << i);
mort = (mort |= xbit)<<i+1; /*shifting*/
//copy formed code to copy ,so that next time the value is preserved for appending
x_y_copy|= mort;
mort =0;
//take 32nd bit from 'y' also
int ybit = ((unsigned long)y)& (1 << i);
mort = (mort |= ybit)<<i;
x_y_copy |= mort;
}
//this is important,when 'y' is negative because the 32nd bit of 'y' is set to 0 by above first code,and while moving 32 bit of 'y' to morton code,the value 0 is copied to 63rd bit,which has to be made to 1,as sign bit is not 63rd bit.
if (mapu_y < 0)
{
x_y_copy = (x_y_copy) | (4611686018427387904);//4611686018427387904 = pow(2,63)
}