geohash的算法
下面以(39.92324, 116.3906)为例，介绍一下geohash的编码算法。首先将纬度范围(-90, 90)平分成两个区间(-90, 0)、(0, 90)， 如果目标纬度位于前一个区间，则编码为0，否则编码为1。由于39.92324属于(0, 90)，所以取编码为1。然后再将(0, 90)分成 (0, 45), (45, 90)两个区间，而39.92324位于(0, 45)，所以编码为0。以此类推，直到精度符合要求为止，得到纬度编码为1011 1000 1100 0111 1001。
| 纬度范围 | 划分区间0 | 划分区间1 | 39.92324所属区间 |
| (-90, 90) | (-90, 0.0) | (0.0, 90) | 1 |
| (0.0, 90) | (0.0, 45.0) | (45.0, 90) | 0 |
| (0.0, 45.0) | (0.0, 22.5) | (22.5, 45.0) | 1 |
| (22.5, 45.0) | (22.5, 33.75) | (33.75, 45.0) | 1 |
| (33.75, 45.0) | (33.75, 39.375) | (39.375, 45.0) | 1 |
| (39.375, 45.0) | (39.375, 42.1875) | (42.1875, 45.0) | 0 |
| (39.375, 42.1875) | (39.375, 40.7812) | (40.7812, 42.1875) | 0 |
| (39.375, 40.7812) | (39.375, 40.0781) | (40.0781, 40.7812) | 0 |
| (39.375, 40.0781) | (39.375, 39.7265) | (39.7265, 40.0781) | 1 |
| (39.7265, 40.0781) | (39.7265, 39.9023) | (39.9023, 40.0781) | 1 |
| (39.9023, 40.0781) | (39.9023, 39.9902) | (39.9902, 40.0781) | 0 |
| (39.9023, 39.9902) | (39.9023, 39.9462) | (39.9462, 39.9902) | 0 |
| (39.9023, 39.9462) | (39.9023, 39.9243) | (39.9243, 39.9462) | 0 |
| (39.9023, 39.9243) | (39.9023, 39.9133) | (39.9133, 39.9243) | 1 |
| (39.9133, 39.9243) | (39.9133, 39.9188) | (39.9188, 39.9243) | 1 |
| (39.9188, 39.9243) | (39.9188, 39.9215) | (39.9215, 39.9243) | 1 |
经度也用同样的算法，对(-180, 180)依次细分，得到116.3906的编码为1101 0010 1100 0100 0100。
| 经度范围 | 划分区间0 | 划分区间1 | 116.3906所属区间 |
| (-180, 180) | (-180, 0.0) | (0.0, 180) | 1 |
| (0.0, 180) | (0.0, 90.0) | (90.0, 180) | 1 |
| (90.0, 180) | (90.0, 135.0) | (135.0, 180) | 0 |
| (90.0, 135.0) | (90.0, 112.5) | (112.5, 135.0) | 1 |
| (112.5, 135.0) | (112.5, 123.75) | (123.75, 135.0) | 0 |
| (112.5, 123.75) | (112.5, 118.125) | (118.125, 123.75) | 0 |
| (112.5, 118.125) | (112.5, 115.312) | (115.312, 118.125) | 1 |
| (115.312, 118.125) | (115.312, 116.718) | (116.718, 118.125) | 0 |
| (115.312, 116.718) | (115.312, 116.015) | (116.015, 116.718) | 1 |
| (116.015, 116.718) | (116.015, 116.367) | (116.367, 116.718) | 1 |
| (116.367, 116.718) | (116.367, 116.542) | (116.542, 116.718) | 0 |
| (116.367, 116.542) | (116.367, 116.455) | (116.455, 116.542) | 0 |
| (116.367, 116.455) | (116.367, 116.411) | (116.411, 116.455) | 0 |
| (116.367, 116.411) | (116.367, 116.389) | (116.389, 116.411) | 1 |
| (116.389, 116.411) | (116.389, 116.400) | (116.400, 116.411) | 0 |
| (116.389, 116.400) | (116.389, 116.394) | (116.394, 116.400) | 0 |
接下来将经度和纬度的编码合并，奇数位是纬度，偶数位是经度，得到编码 11100 11101 00100 01111 00000 01101 01011 00001。
最后，用0-9、b-z（去掉a, i, l, o）这32个字母进行base32编码，得到(39.92324, 116.3906)的编码为wx4g0ec1。
| 十进制 | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 |
| base32 | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | b | c | d | e | f | g |
| 十进制 | 16 | 17 | 18 | 19 | 20 | 21 | 22 | 23 | 24 | 25 | 26 | 27 | 28 | 29 | 30 | 31 |
| base32 | h | j | k | m | n | p | q | r | s | t | u | v | w | x | y | z |
解码算法与编码算法相反，先进行base32解码，然后分离出经纬度，最后根据二进制编码对经纬度范围进行细分即可，这里不再赘述。 不过由于geohash表示的是区间，编码越长越精确，但不可能解码出完全一致的地址。
#define BASE32 "0123456789bcdefghjkmnpqrstuvwxyz" static void encode_geohash(double latitude, double longitude, int precision, char *geohash) { int is_even=1, i=0; double lat[2], lon[2], mid; char bits[] = {16,8,4,2,1}; int bit=0, ch=0; lat[0] = -90.0; lat[1] = 90.0; lon[0] = -180.0; lon[1] = 180.0; while (i < precision) { if (is_even) { mid = (lon[0] + lon[1]) / 2; if (longitude > mid) { ch |= bits[bit]; lon[0] = mid; } else lon[1] = mid; } else { mid = (lat[0] + lat[1]) / 2; if (latitude > mid) { ch |= bits[bit]; lat[0] = mid; } else lat[1] = mid; } is_even = !is_even; if (bit < 4) bit++; else { geohash[i++] = BASE32[ch]; bit = 0; ch = 0; } } geohash[i] = 0; }