如何使用 C++ 将字符串散列到 int？
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How can I hash a string to an int using c++?
提问by zebraman
I have to write my own hash function. If I wanted to just make the simple hash function that maps each letter in the string to a numerical value (i.e. a=1, b=2, c=3, ...), is there a way I can perform this hash on a string without having to first convert it to a c-string to look at each individual char? Is there a more efficient way of hashing strings?
我必须编写自己的哈希函数。如果我只想制作将字符串中的每个字母映射到数值（即 a=1, b=2, c=3, ...）的简单散列函数，有没有一种方法可以在一个字符串而不必先将其转换为 c 字符串来查看每个单独的字符？有没有更有效的方法来散列字符串？
采纳答案by Alex Martelli
Re the first question, sure, e.g, something like:
关于第一个问题，当然，例如，类似于：
int hash = 0;
int offset = 'a' - 1;
for(string::const_iterator it=s.begin(); it!=s.end(); ++it) {
hash = hash << 1 | (*it - offset);
}
regarding the second, there are many better ways to hash strings. E.g., see herefor a few C examples (easily translatable to C++ along the lines of the snippet above).
关于第二个，有很多更好的方法来散列字符串。例如，请参阅此处了解一些 C 示例（可以按照上面的代码段轻松翻译为 C++）。
回答by Tim Cooper
From personal experience I know that this works and produces good distributions. (Plagiarised from http://www.cse.yorku.ca/~oz/hash.html):
根据个人经验，我知道这有效并产生了良好的发行版。（抄袭自http://www.cse.yorku.ca/~oz/hash.html）：
djb2
djb2
this algorithm (k=33) was first reported by dan bernstein many years ago in comp.lang.c. another version of this algorithm (now favored by bernstein) uses xor: hash(i) = hash(i - 1) * 33 ^ str[i]; the magic of number 33 (why it works better than many other constants, prime or not) has never been adequately explained.
该算法 (k=33) 多年前由 dan Bernstein 在 comp.lang.c 中首次报道。该算法的另一个版本（现在受到伯恩斯坦青睐）使用异或： hash(i) = hash(i - 1) * 33 ^ str[i]; 数字 33 的魔力（为什么它比许多其他常数更有效，无论质数与否）从未得到充分解释。
unsigned long hash(unsigned char *str) {
unsigned long hash = 5381;
int c;
while (c = *str++) {
hash = ((hash << 5) + hash) + c; /* hash * 33 + c */
}
return hash;
}
回答by wheaties
You can examine each individual char from a std::string using the []operator. However, you can look at Boost::Functional/Hashfor guidance on a better hashing scheme. There is also a list of hashing functions in c located here.
您可以使用[]运算符检查 std::string 中的每个单独的字符。但是，您可以查看Boost::Functional/Hash以获得关于更好的散列方案的指导。此处还有一个 c 中的散列函数列表。
回答by Wren
Here's a C (++) hash function that I found in Stroustrup's book:
这是我在 Stroustrup 的书中找到的 C (++) 哈希函数：
int hash(const char *str)
{
int h = 0;
while (*str)
h = h << 1 ^ *str++;
return h;
}
If you're using it for a hash table (which Stroustrup does) then you can instead return the abs of the hash modulo a prime number. So instead
如果您将它用于哈希表（Stroustrup 这样做），那么您可以改为返回哈希模的 abs 一个素数。所以与其
return (h > 0 ? h : -h) % N_BUCKETS;
for the last line.
对于最后一行。
回答by alfC
C++11 ships with a standard hashing function for strings.
C++11 附带了一个标准的字符串散列函数。
https://en.cppreference.com/w/cpp/string/basic_string/hash
https://en.cppreference.com/w/cpp/string/basic_string/hash
#include <string>
#include<functional> // hash
int main(){
std::string s = "Hello";
std::size_t hash = std::hash<std::string>{}(s);
}
回答by Stephen
xor the characters together, four at a time.
将字符异或在一起，一次四个。
回答by LUCAS
Another way for small strings:
小字符串的另一种方法：
int hash(const char* str) {
int hash = 0;
int c = 0;
while (c < std::strlen(str)) {
hash += (int)str[c] << (int)str[c+1];
c++;
}
return hash;
}
回答by wilhelmtell
#include <iostream>
#include <string>
#include <algorithm>
using namespace std;
// a variation on dan bernstein's algorithm
// [http://www.cse.yorku.ca/~oz/hash.html]
template<typename Int>
struct hash {
hash() : acc(5381) { }
template<typename Ch>
void operator()(Ch ch) { acc = ((acc << 5) + acc) ^ ch; }
operator Int() const { return acc; }
Int acc;
};
int main(int argc, char* argv[])
{
string s("Hellp, world");
cout << hex << showbase
<< for_each(s.begin(), s.end(), hash<unsigned long long>()) << '\n';
return 0;
}
回答by codaddict
You can make use of the member functions operator[]or atof the string class or iterators to access individual char of a string object without converting it to c-style char array.
您可以使用成员函数operator[]或字符串类或迭代器的at来访问字符串对象的单个字符，而无需将其转换为 c 样式的字符数组。
To hash a string object to an integer you'll have to access each individual char of the string object which you can do as:
要将字符串对象散列为整数，您必须访问字符串对象的每个单独字符，您可以这样做：
for (i=0; i < str.length(); i++) {
// use str[i] or str.at(i) to access ith element.
}
