【问题标题】:Where can I find source or algorithm of Python's hash() function?我在哪里可以找到 Python 的 hash() 函数的源代码或算法?
【发布时间】:2011-01-05 10:20:15
【问题描述】:
>>> hash("\x01")
128000384
>>> hash("\x02")
256000771
>>> hash("\x03")
384001154
>>> hash("\x04")
512001541

有趣的部分是128000384 x 2 不是256000771,还有其他的

我只是想知道该算法是如何工作的,并想从中学习一些东西。

【问题讨论】:

  • (128000384 * 2 != 256000771) 有什么有趣的地方?你知道 (2 * "\x01" != "\x02") 吗?
  • 好吧,在我看到那些哈希值之前我并没有意识到,但是 128000.. 和 256000... 让我觉得其中有一些关系。
  • 请注意,您不应依赖hash() 的任何特定行为。它可能因版本而异,对于某些对象,甚至因运行而异。
  • hash 可以因版本而异,但每次运行都应该保持一致,否则hash函数和随机函数没有区别

标签: python c algorithm hash


【解决方案1】:

如果你下载了 Python 的源代码,你一定会找到的! 但请记住,散列函数对每种对象的实现方式不同。

例如,您会在函数unicode_hash 中找到Objects/unicodeobject.c 中的unicode 哈希函数。您可能需要多看一点才能找到字符串散列函数。找到定义您感兴趣的对象的结构,并在字段tp_hash 中找到计算该对象哈希码的函数。

对于字符串对象:具体代码在函数string_hash中的Objects/stringobject.c中找到:

static long string_hash(PyStringObject *a)
{
    register Py_ssize_t len;
    register unsigned char *p;
    register long x;

    if (a->ob_shash != -1)
        return a->ob_shash;
    len = Py_SIZE(a);
    p = (unsigned char *) a->ob_sval;
    x = *p << 7;
    while (--len >= 0)
        x = (1000003*x) ^ *p++;
    x ^= Py_SIZE(a);
    if (x == -1)
        x = -2;
    a->ob_shash = x;
    return x;
}

【讨论】:

  • 谢谢,我有源代码,但是哈希这个词有很多命中,我找不到,还是谢谢。 +1
  • 我编辑了一点,为您提供有关如何找到您感兴趣的哈希的更多信息。
  • 非常感谢,我现在也发现了。
  • 这是long 32 位还是64 位?还是在不同的系统上有所不同?
  • C 标准未定义有符号整数的溢出,所以乘法是否会遇到未定义的行为?
【解决方案2】:

我认为接受的答案不能真正代表cPython的内部哈希实现,可以在pyhash.c找到:

数值类型的散列算法说明:

/* For numeric types, the hash of a number x is based on the reduction
   of x modulo the prime P = 2**_PyHASH_BITS - 1.  It's designed so that
   hash(x) == hash(y) whenever x and y are numerically equal, even if
   x and y have different types.

   A quick summary of the hashing strategy:

   (1) First define the 'reduction of x modulo P' for any rational
   number x; this is a standard extension of the usual notion of
   reduction modulo P for integers.  If x == p/q (written in lowest
   terms), the reduction is interpreted as the reduction of p times
   the inverse of the reduction of q, all modulo P; if q is exactly
   divisible by P then define the reduction to be infinity.  So we've
   got a well-defined map

      reduce : { rational numbers } -> { 0, 1, 2, ..., P-1, infinity }.

   (2) Now for a rational number x, define hash(x) by:

      reduce(x)   if x >= 0
      -reduce(-x) if x < 0

   If the result of the reduction is infinity (this is impossible for
   integers, floats and Decimals) then use the predefined hash value
   _PyHASH_INF for x >= 0, or -_PyHASH_INF for x < 0, instead.
   _PyHASH_INF and -_PyHASH_INF are also used for the
   hashes of float and Decimal infinities.

   NaNs hash with a pointer hash.  Having distinct hash values prevents
   catastrophic pileups from distinct NaN instances which used to always
   have the same hash value but would compare unequal.

   A selling point for the above strategy is that it makes it possible
   to compute hashes of decimal and binary floating-point numbers
   efficiently, even if the exponent of the binary or decimal number
   is large.  The key point is that

      reduce(x * y) == reduce(x) * reduce(y) (modulo _PyHASH_MODULUS)

   provided that {reduce(x), reduce(y)} != {0, infinity}.  The reduction of a
   binary or decimal float is never infinity, since the denominator is a power
   of 2 (for binary) or a divisor of a power of 10 (for decimal).  So we have,
   for nonnegative x,

      reduce(x * 2**e) == reduce(x) * reduce(2**e) % _PyHASH_MODULUS

      reduce(x * 10**e) == reduce(x) * reduce(10**e) % _PyHASH_MODULUS

   and reduce(10**e) can be computed efficiently by the usual modular
   exponentiation algorithm.  For reduce(2**e) it's even better: since
   P is of the form 2**n-1, reduce(2**e) is 2**(e mod n), and multiplication
   by 2**(e mod n) modulo 2**n-1 just amounts to a rotation of bits.

   */

双打散列:

Py_hash_t
_Py_HashDouble(double v)
{
    int e, sign;
    double m;
    Py_uhash_t x, y;

    if (!Py_IS_FINITE(v)) {
        if (Py_IS_INFINITY(v))
            return v > 0 ? _PyHASH_INF : -_PyHASH_INF;
        else
            return _PyHASH_NAN;
    }

    m = frexp(v, &e);

    sign = 1;
    if (m < 0) {
        sign = -1;
        m = -m;
    }

    /* process 28 bits at a time;  this should work well both for binary
       and hexadecimal floating point. */
    x = 0;
    while (m) {
        x = ((x << 28) & _PyHASH_MODULUS) | x >> (_PyHASH_BITS - 28);
        m *= 268435456.0;  /* 2**28 */
        e -= 28;
        y = (Py_uhash_t)m;  /* pull out integer part */
        m -= y;
        x += y;
        if (x >= _PyHASH_MODULUS)
            x -= _PyHASH_MODULUS;
    }

    /* adjust for the exponent;  first reduce it modulo _PyHASH_BITS */
    e = e >= 0 ? e % _PyHASH_BITS : _PyHASH_BITS-1-((-1-e) % _PyHASH_BITS);
    x = ((x << e) & _PyHASH_MODULUS) | x >> (_PyHASH_BITS - e);

    x = x * sign;
    if (x == (Py_uhash_t)-1)
        x = (Py_uhash_t)-2;
    return (Py_hash_t)x;
}

指针散列:

Py_hash_t
_Py_HashPointerRaw(const void *p)
{
    size_t y = (size_t)p;
    /* bottom 3 or 4 bits are likely to be 0; rotate y by 4 to avoid
       excessive hash collisions for dicts and sets */
    y = (y >> 4) | (y << (8 * SIZEOF_VOID_P - 4));
    return (Py_hash_t)y;
}

Py_hash_t
_Py_HashPointer(const void *p)
{
    Py_hash_t x = _Py_HashPointerRaw(p);
    if (x == -1) {
        x = -2;
    }
    return x;
}

非常短字符串的字节散列使用 DJBX33A,否则使用默认散列:

Py_hash_t
_Py_HashBytes(const void *src, Py_ssize_t len)
{
    Py_hash_t x;
    /*
      We make the hash of the empty string be 0, rather than using
      (prefix ^ suffix), since this slightly obfuscates the hash secret
    */
    if (len == 0) {
        return 0;
    }

#ifdef Py_HASH_STATS
    hashstats[(len <= Py_HASH_STATS_MAX) ? len : 0]++;
#endif

#if Py_HASH_CUTOFF > 0
    if (len < Py_HASH_CUTOFF) {
        /* Optimize hashing of very small strings with inline DJBX33A. */
        Py_uhash_t hash;
        const unsigned char *p = src;
        hash = 5381; /* DJBX33A starts with 5381 */

        switch(len) {
            /* ((hash << 5) + hash) + *p == hash * 33 + *p */
            case 7: hash = ((hash << 5) + hash) + *p++; /* fallthrough */
            case 6: hash = ((hash << 5) + hash) + *p++; /* fallthrough */
            case 5: hash = ((hash << 5) + hash) + *p++; /* fallthrough */
            case 4: hash = ((hash << 5) + hash) + *p++; /* fallthrough */
            case 3: hash = ((hash << 5) + hash) + *p++; /* fallthrough */
            case 2: hash = ((hash << 5) + hash) + *p++; /* fallthrough */
            case 1: hash = ((hash << 5) + hash) + *p++; break;
            default:
                Py_UNREACHABLE();
        }
        hash ^= len;
        hash ^= (Py_uhash_t) _Py_HashSecret.djbx33a.suffix;
        x = (Py_hash_t)hash;
    }
    else
#endif /* Py_HASH_CUTOFF */
        x = PyHash_Func.hash(src, len);

    if (x == -1)
        return -2;
    return x;
}

该文件还实现了修改后的 FNV 哈希:

#if Py_HASH_ALGORITHM == Py_HASH_FNV
/* **************************************************************************
 * Modified Fowler-Noll-Vo (FNV) hash function
 */
static Py_hash_t
fnv(const void *src, Py_ssize_t len)
{
    const unsigned char *p = src;
    Py_uhash_t x;
    Py_ssize_t remainder, blocks;
    union {
        Py_uhash_t value;
        unsigned char bytes[SIZEOF_PY_UHASH_T];
    } block;

#ifdef Py_DEBUG
    assert(_Py_HashSecret_Initialized);
#endif
    remainder = len % SIZEOF_PY_UHASH_T;
    if (remainder == 0) {
        /* Process at least one block byte by byte to reduce hash collisions
         * for strings with common prefixes. */
        remainder = SIZEOF_PY_UHASH_T;
    }
    blocks = (len - remainder) / SIZEOF_PY_UHASH_T;

    x = (Py_uhash_t) _Py_HashSecret.fnv.prefix;
    x ^= (Py_uhash_t) *p << 7;
    while (blocks--) {
        PY_UHASH_CPY(block.bytes, p);
        x = (_PyHASH_MULTIPLIER * x) ^ block.value;
        p += SIZEOF_PY_UHASH_T;
    }
    /* add remainder */
    for (; remainder > 0; remainder--)
        x = (_PyHASH_MULTIPLIER * x) ^ (Py_uhash_t) *p++;
    x ^= (Py_uhash_t) len;
    x ^= (Py_uhash_t) _Py_HashSecret.fnv.suffix;
    if (x == -1) {
        x = -2;
    }
    return x;
}

static PyHash_FuncDef PyHash_Func = {fnv, "fnv", 8 * SIZEOF_PY_HASH_T,
                                     16 * SIZEOF_PY_HASH_T};

#endif /* Py_HASH_ALGORITHM == Py_HASH_FNV */

根据PEP 456,SipHash(MIT License)是默认的字符串和字节哈希算法:

/* byte swap little endian to host endian
 * Endian conversion not only ensures that the hash function returns the same
 * value on all platforms. It is also required to for a good dispersion of
 * the hash values' least significant bits.
 */
#if PY_LITTLE_ENDIAN
#  define _le64toh(x) ((uint64_t)(x))
#elif defined(__APPLE__)
#  define _le64toh(x) OSSwapLittleToHostInt64(x)
#elif defined(HAVE_LETOH64)
#  define _le64toh(x) le64toh(x)
#else
#  define _le64toh(x) (((uint64_t)(x) << 56) | \
                      (((uint64_t)(x) << 40) & 0xff000000000000ULL) | \
                      (((uint64_t)(x) << 24) & 0xff0000000000ULL) | \
                      (((uint64_t)(x) << 8)  & 0xff00000000ULL) | \
                      (((uint64_t)(x) >> 8)  & 0xff000000ULL) | \
                      (((uint64_t)(x) >> 24) & 0xff0000ULL) | \
                      (((uint64_t)(x) >> 40) & 0xff00ULL) | \
                      ((uint64_t)(x)  >> 56))
#endif


#ifdef _MSC_VER
#  define ROTATE(x, b)  _rotl64(x, b)
#else
#  define ROTATE(x, b) (uint64_t)( ((x) << (b)) | ( (x) >> (64 - (b))) )
#endif

#define HALF_ROUND(a,b,c,d,s,t)         \
    a += b; c += d;             \
    b = ROTATE(b, s) ^ a;           \
    d = ROTATE(d, t) ^ c;           \
    a = ROTATE(a, 32);

#define DOUBLE_ROUND(v0,v1,v2,v3)       \
    HALF_ROUND(v0,v1,v2,v3,13,16);      \
    HALF_ROUND(v2,v1,v0,v3,17,21);      \
    HALF_ROUND(v0,v1,v2,v3,13,16);      \
    HALF_ROUND(v2,v1,v0,v3,17,21);


static uint64_t
siphash24(uint64_t k0, uint64_t k1, const void *src, Py_ssize_t src_sz) {
    uint64_t b = (uint64_t)src_sz << 56;
    const uint64_t *in = (uint64_t*)src;

    uint64_t v0 = k0 ^ 0x736f6d6570736575ULL;
    uint64_t v1 = k1 ^ 0x646f72616e646f6dULL;
    uint64_t v2 = k0 ^ 0x6c7967656e657261ULL;
    uint64_t v3 = k1 ^ 0x7465646279746573ULL;

    uint64_t t;
    uint8_t *pt;
    uint8_t *m;

    while (src_sz >= 8) {
        uint64_t mi = _le64toh(*in);
        in += 1;
        src_sz -= 8;
        v3 ^= mi;
        DOUBLE_ROUND(v0,v1,v2,v3);
        v0 ^= mi;
    }

    t = 0;
    pt = (uint8_t *)&t;
    m = (uint8_t *)in;
    switch (src_sz) {
        case 7: pt[6] = m[6]; /* fall through */
        case 6: pt[5] = m[5]; /* fall through */
        case 5: pt[4] = m[4]; /* fall through */
        case 4: memcpy(pt, m, sizeof(uint32_t)); break;
        case 3: pt[2] = m[2]; /* fall through */
        case 2: pt[1] = m[1]; /* fall through */
        case 1: pt[0] = m[0]; /* fall through */
    }
    b |= _le64toh(t);

    v3 ^= b;
    DOUBLE_ROUND(v0,v1,v2,v3);
    v0 ^= b;
    v2 ^= 0xff;
    DOUBLE_ROUND(v0,v1,v2,v3);
    DOUBLE_ROUND(v0,v1,v2,v3);

    /* modified */
    t = (v0 ^ v1) ^ (v2 ^ v3);
    return t;
}

static Py_hash_t
pysiphash(const void *src, Py_ssize_t src_sz) {
    return (Py_hash_t)siphash24(
        _le64toh(_Py_HashSecret.siphash.k0), _le64toh(_Py_HashSecret.siphash.k1),
        src, src_sz);
}

uint64_t
_Py_KeyedHash(uint64_t key, const void *src, Py_ssize_t src_sz)
{
    return siphash24(key, 0, src, src_sz);
}


#if Py_HASH_ALGORITHM == Py_HASH_SIPHASH24
static PyHash_FuncDef PyHash_Func = {pysiphash, "siphash24", 64, 128};
#endif

(tupleobject.c) 中的元组等对象有自己的哈希方法。有关更多示例,请参见源代码:

static Py_hash_t
tuplehash(PyTupleObject *v)
{
    Py_uhash_t x;  /* Unsigned for defined overflow behavior. */
    Py_hash_t y;
    Py_ssize_t len = Py_SIZE(v);
    PyObject **p;
    Py_uhash_t mult = _PyHASH_MULTIPLIER;
    x = 0x345678UL;
    p = v->ob_item;
    while (--len >= 0) {
        y = PyObject_Hash(*p++);
        if (y == -1)
            return -1;
        x = (x ^ y) * mult;
        /* the cast might truncate len; that doesn't change hash stability */
        mult += (Py_hash_t)(82520UL + len + len);
    }
    x += 97531UL;
    if (x == (Py_uhash_t)-1)
        x = -2;
    return x;
}

【讨论】:

  • 这个答案很完美。
  • 你知道是否有一种方法可以直接调用siphash24(指定k0k1,而不是_Py_HashSecret 中的那些使哈希不同于一次Python 调用下一个)?
  • @PierreD 我不太了解 python 内部如何组合在一起。作为一个单独的问题可能更好。这里的几乎所有代码都是自包含的,因此您可以将其复制到自己的文件中并运行它。
  • @qwr 确实是这样,但这会使我的 conda 构建复杂化,我必须专门为每个拱门发布。由于它“已经存在于 Python 中”,因此很容易找到一种直接使用它而不添加更多包依赖项或构建复杂性的方法。
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