c - 二维莫顿解码功能 64bits

Question

第一个函数将 [x, y] 编码为 64 位宽的 Morton 代码，其中 x 和 y 是 32 位宽的整数，使用二进制魔数(Magic Number)的交错位。

反向函数是什么？

void xy2d_morton_64bits(uint64_t x, uint64_t y, uint64_t *d)
{
    x = (x | (x << 16)) & 0x0000FFFF0000FFFF;
    x = (x | (x << 8)) & 0x00FF00FF00FF00FF;   
    x = (x | (x << 4)) & 0x0F0F0F0F0F0F0F0F; 
    x = (x | (x << 2)) & 0x3333333333333333;
    x = (x | (x << 1)) & 0x5555555555555555;

    y = (y | (y << 16)) & 0x0000FFFF0000FFFF;
    y = (y | (y << 8)) & 0x00FF00FF00FF00FF;
    y = (y | (y << 4)) & 0x0F0F0F0F0F0F0F0F;
    y = (y | (y << 2)) & 0x3333333333333333;
    y = (y | (y << 1)) & 0x5555555555555555;

    *d = x | (y << 1);
}

void d2xy_morton_64bits(uint64_t d, uint64_t *x, uint64_t *y)
{
    // ????
}

Answer 1

void xy2d_morton(uint64_t x, uint64_t y, uint64_t *d)
{
    x = (x | (x << 16)) & 0x0000FFFF0000FFFF;
    x = (x | (x << 8)) & 0x00FF00FF00FF00FF;
    x = (x | (x << 4)) & 0x0F0F0F0F0F0F0F0F;
    x = (x | (x << 2)) & 0x3333333333333333;
    x = (x | (x << 1)) & 0x5555555555555555;

    y = (y | (y << 16)) & 0x0000FFFF0000FFFF;
    y = (y | (y << 8)) & 0x00FF00FF00FF00FF;
    y = (y | (y << 4)) & 0x0F0F0F0F0F0F0F0F;
    y = (y | (y << 2)) & 0x3333333333333333;
    y = (y | (y << 1)) & 0x5555555555555555;

    *d = x | (y << 1);
}

// morton_1 - extract even bits

uint64_t morton_1(uint64_t x)
{
    x = x & 0x5555555555555555;
    x = (x | (x >> 1)) & 0x3333333333333333;
    x = (x | (x >> 2)) & 0x0F0F0F0F0F0F0F0F;
    x = (x | (x >> 4)) & 0x00FF00FF00FF00FF;
    x = (x | (x >> 8)) & 0x0000FFFF0000FFFF;
    x = (x | (x >> 16)) & 0xFFFFFFFFFFFFFFFF;
    return x;
}

void d2xy_morton(uint64_t d, uint64_t *x, uint64_t *y)
{
    *x = morton_1(d);
    *y = morton_1(d >> 1);
}