小编典典

如何在Python 3.2中实现__hash__?

algorithm

我想使自定义对象可哈希化(通过腌制)。我可以找到__hash__适用于Python 2.x的算法(请参见下面的代码),但是它显然 不同于
适用于Python 3.2的哈希(我想知道为什么吗?)。有人知道如何__hash__在Python 3.2中实现吗?

#Version: Python 3.2

def c_mul(a, b):
    #C type multiplication
    return eval(hex((int(a) * b) & 0xFFFFFFFF)[:-1])

class hs:
    #Python 2.x algorithm for hash from http://effbot.org/zone/python-hash.htm
    def __hash__(self):
        if not self:
            return 0 # empty
        value = ord(self[0]) << 7
        for char in self:
            value = c_mul(1000003, value) ^ ord(char)
        value = value ^ len(self)
        if value == -1:
            value = -2
        return value


def main():
    s = ["PROBLEM", "PROBLEN", "PROBLEO", "PROBLEP"]#, "PROBLEQ", "PROBLER", "PROBLES"]
    print("Python 3.2 hash() bild-in")
    for c in s[:]: print("hash('", c, "')=", hex(hash(c)),  end="\n")
    print("\n")
    print("Python 2.x type hash: __hash__()")
    for c in s[:]: print("hs.__hash__('", c, "')=", hex(hs.__hash__(c)),  end="\n")


if __name__ == "__main__":
    main()

OUTPUT:
Python 3.2 hash() bild-in
hash(' PROBLEM ')= 0x7a8e675a
hash(' PROBLEN ')= 0x7a8e6759
hash(' PROBLEO ')= 0x7a8e6758
hash(' PROBLEP ')= 0x7a8e6747


Python 2.x type hash: __hash__()
hs.__hash__(' PROBLEM ')= 0xa638a41
hs.__hash__(' PROBLEN ')= 0xa638a42
hs.__hash__(' PROBLEO ')= 0xa638a43
hs.__hash__(' PROBLEP ')= 0xa638a5c

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2020-07-28

共1个答案

小编典典

它们为何不同的答案写在这里:

哈希值现在是一种新类型的值Py_hash_t,该值定义为与指针大小相同。以前它们是long型的,在某些64位操作系统上,它们仍然只有32位长。

哈希还考虑了要计算的新值,看看

 sys.hash_info

对于字符串,您可以查看http://svn.python.org/view/python/trunk/Objects/stringobject.c?view=markup第1263行string_hash(PyStringObject * a)

2020-07-28