blocks|key|1512449|text|您可以使用[]操作符检查std::string中的每个字符。但是，您可以查看Boost::Functional/Hash以获得更好的散列方案的指导。在c中的here中也有一个散列函数列表。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1512450|entityMap|0|LINK|mutability|MUTABLE|url|http://www.boost.org/doc/libs/1_37_0/doc/html/hash.html|1|http://www.cse.yorku.ca/~oz/hash.html^0|5|2|12|M|0|26|4|1|0^^$0|@$1|2|3|4|5|6|7|P|8|@$9|Q|A|R|B|C]]|D|@$9|S|A|T|1|U]|$9|V|A|W|1|X]]|E|$]]|$1|F|3|-4|5|6|7|Y|8|@]|D|@]|E|$]]]|G|$H|$5|I|J|K|E|$L|M]]|N|$5|I|J|K|E|$L|O]]]]

You can examine each individual char from a std::string using the <code>[]</code> operator. However, you can look at <a href="http://www.boost.org/doc/libs/1_37_0/doc/html/hash.html" rel="noreferrer">Boost::Functional/Hash</a> for guidance on a better hashing scheme. There is also a list of hashing functions in c located <a href="http://www.cse.yorku.ca/~oz/hash.html" rel="noreferrer">here</a>.

blocks|key|1512485|text|您可以使用string类的成员函数operator[]或at或迭代器来访问string对象的单个字符，而无需将其转换为c样式的字符数组。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1512486|要将string对象散列为一个整数，您必须访问string对象的每个字符，您可以这样做：|1512487|for+(i=0;+i+<+str.length();+i%2B%2B)+{
++++//+use+str[i]+or+str.at(i)+to+access+ith+element.
}|code-block|syntax|javascript|1512488|entityMap|0|LINK|mutability|MUTABLE|url|http://www.cplusplus.com/reference/string/string/operator%255B%255D/|1|http://www.cplusplus.com/reference/string/string/at/^0|H|A|0|S|2|1|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]|$A|Y|B|Z|1|10]]|C|$]]|$1|D|3|E|5|6|7|11|8|@]|9|@]|C|$]]|$1|F|3|G|5|H|7|12|8|@]|9|@]|C|$I|J]]|$1|K|3|-4|5|6|7|13|8|@]|9|@]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]|S|$5|N|O|P|C|$Q|T]]]]

You can make use of the member functions <a href="http://www.cplusplus.com/reference/string/string/operator%5B%5D/" rel="nofollow noreferrer">operator[]</a> or <a href="http://www.cplusplus.com/reference/string/string/at/" rel="nofollow noreferrer">at</a> of the string class or iterators to access individual char of a string object without converting it to c-style char array.

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:

<pre><code>for (i=0; i &lt; str.length(); i++) {
 // use str[i] or str.at(i) to access ith element.
}
</code></pre>

blocks|key|1024020|text|对字符进行xor运算，一次四个。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1024021|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

xor the characters together, four at a time.

blocks|key|1020301|text|你是第一个问题，当然，例如，就像这样：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1020302|int+hash+=+0;
int+offset+=+'a'+-+1;
for(string::const_iterator+it=s.begin();+it!=s.end();+%2B%2Bit)+{
++hash+=+hash+<<+1+%7C+(*it+-+offset);
}|code-block|syntax|javascript|1020303|关于第二种方法，有很多更好的方法来散列字符串。例如，参见here中的几个C示例(很容易按照上面代码片段的行将其翻译成C%2B%2B+)。|offset|length|1020304|entityMap|0|LINK|mutability|MUTABLE|url|http://www.cse.yorku.ca/~oz/hash.html^0|0|0|S|4|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@$I|V|J|W|1|X]]|A|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

Re the first question, sure, e.g, something like:

<pre><code>int hash = 0;
int offset = 'a' - 1;
for(string::const_iterator it=s.begin(); it!=s.end(); ++it) {
 hash = hash &lt;&lt; 1 | (*it - offset);
}
</code></pre>

regarding the second, there are many better ways to hash strings. E.g., see <a href="http://www.cse.yorku.ca/~oz/hash.html" rel="nofollow noreferrer">here</a> for a few C examples (easily translatable to C++ along the lines of the snippet above).

blocks|key|2921433|text|#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)+%2B+acc)+%5E+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;
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2921434|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>#include &lt;iostream&gt;
#include &lt;string&gt;
#include &lt;algorithm&gt;

using namespace std;

// a variation on dan bernstein's algorithm
// [http://www.cse.yorku.ca/~oz/hash.html]
template&lt;typename Int&gt;
struct hash {
 hash() : acc(5381) { }
 template&lt;typename Ch&gt;
 void operator()(Ch ch) { acc = ((acc &lt;&lt; 5) + acc) ^ ch; }
 operator Int() const { return acc; }
 Int acc;
};

int main(int argc, char* argv[])
{
 string s("Hellp, world");
 cout &lt;&lt; hex &lt;&lt; showbase
 &lt;&lt; for_each(s.begin(), s.end(), hash&lt;unsigned long long&gt;()) &lt;&lt; '\n';
 return 0;
}
</code></pre>

blocks|key|276203|text|下面是我在Stroustrup的书中找到的一个C+(%2B%2B)散列函数：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|276204|int+hash(const+char+*str)
{
++++int+h+=+0;
++++while+(*str)
+++++++h+=+h+<<+1+%5E+*str%2B%2B;
++++return+h;
}|code-block|syntax|javascript|276205|如果您将其用于哈希表(+Stroustrup就是这么做的)，那么您可以改为返回以素数为模的哈希表的abs。所以相反的是|276206|++++return+(h+>+0+?+h+:+-h)+%25+N_BUCKETS;|276207|最后一行。|276208|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Here's a C (++) hash function that I found in Stroustrup's book:

<pre><code>int hash(const char *str)
{
 int h = 0;
 while (*str)
 h = h &lt;&lt; 1 ^ *str++;
 return h;
}
</code></pre>

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

<pre><code> return (h &gt; 0 ? h : -h) % N_BUCKETS;
</code></pre>

for the last line.

blocks|key|275138|text|根据我的个人经验，我知道这是可行的，并产生了很好的发行版。(抄袭自http://www.cse.yorku.ca/~oz/hash.html)：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|275139|djb2|275140|该算法(k=33)由dan+bernstein于多年前在comp.lang.c上首次报道。该算法的另一个版本(现在得到bernstein的青睐)使用xor:+hash(i)+=+hash(i+-+1)+*+33+%5E+stri；数字33的魔力(为什么它比许多其他常量更好，不管是不是素数)从未得到充分解释。|275141|unsigned+long+hash(unsigned+char+*str)+{
++++unsigned+long+hash+=+5381;
++++int+c;

++++while+(c+=+*str%2B%2B)+{
++++++++hash+=+((hash+<<+5)+%2B+hash)+%2B+c;+/*+hash+*+33+%2B+c+*/
++++}

++++return+hash;
}|code-block|syntax|javascript|275142|entityMap|0|LINK|mutability|MUTABLE|url|http://www.cse.yorku.ca/~oz/hash.html^0|X|11|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|6|7|Y|8|@]|9|@]|C|$]]|$1|F|3|G|5|6|7|Z|8|@]|9|@]|C|$]]|$1|H|3|I|5|J|7|10|8|@]|9|@]|C|$K|L]]|$1|M|3|-4|5|6|7|11|8|@]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]]]

From personal experience I know that this works and produces good distributions. (Plagiarised from <a href="http://www.cse.yorku.ca/~oz/hash.html">http://www.cse.yorku.ca/~oz/hash.html</a>):

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.

<pre><code>unsigned long hash(unsigned char *str) {
 unsigned long hash = 5381;
 int c;

 while (c = *str++) {
 hash = ((hash &lt;&lt; 5) + hash) + c; /* hash * 33 + c */
 }

 return hash;
}
</code></pre>

blocks|key|1512724|text|小字符串的另一种方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1512725|int+hash(const+char*+str)+{
++++int+hash+=+0;
++++int+c+=+0;

++++while+(c+<+std::strlen(str))+{
++++++++hash+%2B=+(int)str[c]+<<+(int)str[c%2B1];
++++++++c%2B%2B;
++++}
++++return+hash;
}|code-block|syntax|javascript|1512726|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Another way for small strings:

<pre><code>int hash(const char* str) {
 int hash = 0;
 int c = 0;

 while (c &lt; std::strlen(str)) {
 hash += (int)str[c] &lt;&lt; (int)str[c+1];
 c++;
 }
 return hash;
}
</code></pre>

blocks|key|276270|text|C%2B%2B11附带了一个标准的字符串散列函数。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|276271|https://en.cppreference.com/w/cpp/string/basic_string/hash|offset|length|276272|#include+<string>
#include<functional>+//+hash
int+main(){
++++std::string+s+=+"Hello";
++++std::size_t+hash+=+std::hash<std::string>{}(s);
}|code-block|syntax|javascript|276273|entityMap|0|LINK|mutability|MUTABLE|url^0|0|0|1M|0|0|0^^$0|@$1|2|3|4|5|6|7|R|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|S|8|@]|9|@$D|T|E|U|1|V]]|A|$]]|$1|F|3|G|5|H|7|W|8|@]|9|@]|A|$I|J]]|$1|K|3|-4|5|6|7|X|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|C]]]]

C++11 ships with a standard hashing function for strings. 

<a href="https://en.cppreference.com/w/cpp/string/basic_string/hash" rel="nofollow noreferrer">https://en.cppreference.com/w/cpp/string/basic_string/hash</a>

<pre><code>#include &lt;string&gt;
#include&lt;functional&gt; // hash
int main(){
 std::string s = "Hello";
 std::size_t hash = std::hash&lt;std::string&gt;{}(s);
}
</code></pre>

blocks|key|1512777|text|刚刚发布了对Arnestig的djb2算法的改进，使其对常量表达式友好。我必须删除参数的无符号限定符，这样它才能与文字字符串一起工作。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1512778|constexpr+unsigned+long+hash(const+char+*str)+{
++++unsigned+long+hash+=+5381;

++++while+(int+c+=+*str%2B%2B)+{
++++++++hash+=+((hash+<<+5)+%2B+hash)+%2B+c;+/*+hash+*+33+%2B+c+*/
++++}

++++return+hash;
}|code-block|syntax|javascript|1512779|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Just posting an improvement to Arnestig's djb2 algorithm to be constexpr-friendly.
I had to remove the unsigned qualifier of the argument so it can work with literal strings.
<pre><code>constexpr unsigned long hash(const char *str) {
 unsigned long hash = 5381;

 while (int c = *str++) {
 hash = ((hash &lt;&lt; 5) + hash) + c; /* hash * 33 + c */
 }

 return hash;
}
</code></pre>

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?

How can I hash a string to an int using c++?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我必须编写自己的散列函数。如果我只想做一个简单的散列函数，将字符串中的每个字母映射到一个数值(即a=1，b=2，c=3，...)，有没有一种方法可以在字符串上执行这种散列，而不必首先将其转换为c字符串来查看每个字符？有没有一种更有效的散列字符串的方法？

问如何使用c++将字符串散列为整型？
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问如何使用c++将字符串散列为整型？EN