c – 哈希表实现

#define NELEMS(x) ((sizeof (x))/(sizeof ((x)[0])))
static struct atom {
    struct atom *link;
    int len;
    char *str;
} *buckets[2048];
static unsigned long scatter[] = {
2078917053,143302914,1027100827,1953210302,755253631,2002600785,1405390230,45248011,1099951567,433832350,2018585307,438263339,813528929,1703199216,618906479,573714703,766270699,275680090,1510320440,1583583926,1723401032,1965443329,1098183682,1636505764,980071615,1011597961,643279273,1315461275,157584038,1069844923,471560540,89017443,1213147837,1498661368,2042227746,1968401469,1353778505,1300134328,2013649480,306246424,1733966678,1884751139,744509763,400011959,1440466707,1363416242,973726663,59253759,1639096332,336563455,1642837685,1215013716,154523136,593537720,704035832,1134594751,1605135681,1347315106,302572379,1762719719,269676381,774132919,1851737163,1482824219,125310639,1746481261,1303742040,1479089144,899131941,1169907872,1785335569,485614972,907175364,382361684,885626931,200158423,1745777927,1859353594,259412182,1237390611,48433401,1902249868,304920680,202956538,348303940,1008956512,1337551289,1953439621,208787970,1640123668,1568675693,478464352,266772940,1272929208,1961288571,392083579,871926821,1117546963,1871172724,1771058762,139971187,1509024645,109190086,1047146551,1891386329,994817018,1247304975,1489680608,706686964,1506717157,579587572,755120366,1261483377,884508252,958076904,1609787317,1893464764,148144545,1415743291,2102252735,1788268214,836935336,433233439,2055041154,2109864544,247038362,299641085,834307717,1364585325,23330161,457882831,1504556512,1532354806,567072918,404219416,1276257488,1561889936,1651524391,618454448,121093252,1010757900,1198042020,876213618,124757630,2082550272,1834290522,1734544947,1828531389,1982435068,1002804590,1783300476,1623219634,1839739926,69050267,1530777140,1802120822,316088629,1830418225,488944891,1680673954,1853748387,946827723,1037746818,1238619545,1513900641,1441966234,367393385,928306929,946006977,985847834,1049400181,1956764878,36406206,1925613800,2081522508,2118956479,1612420674,1668583807,1800004220,1447372094,523904750,1435821048,923108080,216161028,1504871315,306401572,2018281851,1820959944,2136819798,359743094,1354150250,1843084537,1306570817,244413420,934220434,672987810,1686379655,1301613820,1601294739,484902984,139978006,503211273,294184214,176384212,281341425,228223074,147857043,1893762099,1896806882,1947861263,1193650546,273227984,1236198663,2116758626,489389012,593586330,275676551,360187215,267062626,265012701,719930310,1621212876,2108097238,2026501127,1865626297,894834024,552005290,1404522304,48964196,5816381,1889425288,188942202,509027654,36125855,365326415,790369079,264348929,513183458,536647531,13672163,313561074,1730298077,286900147,1549759737,1699573055,776289160,2143346068,1975249606,1136476375,262925046,92778659,1856406685,1884137923,53392249,1735424165,1602280572
};
const char *Atom_new(const char *str,int len) {
    unsigned long h;
    int i;
    struct atom *p;
    assert(str);
    assert(len >= 0);
    for (h = 0,i = 0; i < len; i++)
        h = (h<<1) + scatter[(unsigned char)str[i]];
    h &= NELEMS(buckets)-1;
    for (p = buckets[h]; p; p = p->link)
        if (len == p->len) {
            for (i = 0; i < len && p->str[i] == str[i]; )
                i++;
            if (i == len)
                return p->str;
        }
    p = ALLOC(sizeof (*p) + len + 1);
    p->len = len;
    p->str = (char *)(p + 1);
    if (len > 0)
        memcpy(p->str,str,len);
    p->str[len] = '\0';
    p->link = buckets[h];
    buckets[h] = p;//insert atom in front of list
    return p->str;
}

在本章末尾,在练习3.1中,该书的作者说
“大多数文本都建议使用素数作为
桶.使用素数和良好的散列函数通常会给出一个
更好地分配挂在水桶上的列表的长度.
Atom使用2的幂,有时明确引用
作为一个糟糕的选择.编写一个程序来生成或读取10,000个
典型的字符串和度量Atom_new的速度和分布
列表的长度.然后更换桶以使其具有
2,039个条目(最大素数小于2,048),并重复
测量.使用黄金帮助吗？你有多少
结论取决于你的具体机器？“

所以我确实将哈希表的大小更改为2039,但它似乎是实际制作的素数
列表长度分布不好,我试过64,61,61实际上也做了不好的分配.

我只是想知道为什么一个素数表大小做出错误的分布,这是因为与Atom_new一起使用的哈希函数是一个糟糕的哈希函数吗？

我正在使用此函数打印出原子列表的长度

#define B_SIZE 2048
void Atom_print(void)
{
    int i,t;
    struct atom *atom;
    for(i= 0;i<B_SIZE;i++) {
        t = 0;
        for(atom=buckets[i];atom;atom=atom->link) {
            ++t;
        }
        printf("%d ",t);
    }
}