blocks|key|580491|text|Boost解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|580492|#include+<boost/algorithm/string.hpp>
std::vector<std::string>+strs;
//input_array+must+be+a+Range+containing+the+input.
boost::split(
++++strs,
++++input_array,
++++boost::is_any_of(boost::as_array("\0")));|code-block|syntax|javascript|580493|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|$]]

Boost solution:

<pre><code>#include &lt;boost/algorithm/string.hpp&gt;
std::vector&lt;std::string&gt; strs;
//input_array must be a Range containing the input.
boost::split(
 strs,
 input_array,
 boost::is_any_of(boost::as_array("\0")));
</code></pre>

blocks|key|796395|text|我的两点意见：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|796396|const+char*+p+=+str;
std::vector<std::string>+vector;

do+{
++vector.push_back(std::string(p));
++p+%2B=+vector.back().size()+%2B+1;
}+while+(+//+whatever+condition+applies+);|code-block|syntax|javascript|796397|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|$]]

My two cents :

<pre><code>const char* p = str;
std::vector&lt;std::string&gt; vector;

do {
 vector.push_back(std::string(p));
 p += vector.back().size() + 1;
} while ( // whatever condition applies );
</code></pre>

blocks|key|580557|text|在C中，string.h有这样一个人：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|580558|char+*+strtok+(+char+*+str,+const+char+*+delimiters+);|code-block|syntax|javascript|580559|cplusplus.com上的示例：|580560|/*+strtok+example+*/
#include+<stdio.h>
#include+<string.h>

int+main+()
{
++char+str[]+="-+This,+a+sample+string.";
++char+*+pch;
++printf+("Splitting+string+\"%25s\"+into+tokens:\n",str);
++pch+=+strtok+(str,"+,.-");
++while+(pch+!=+NULL)
++{
++++printf+("%25s\n",pch);
++++pch+=+strtok+(NULL,+"+,.-");
++}
++return+0;
}|580561|它不是C%2B%2B，但它可以工作|580562|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|$]]

In C, string.h has this guy:

<pre><code>char * strtok ( char * str, const char * delimiters );
</code></pre>

the example on cplusplus.com :

<pre><code>/* strtok example */
#include &lt;stdio.h&gt;
#include &lt;string.h&gt;

int main ()
{
 char str[] ="- This, a sample string.";
 char * pch;
 printf ("Splitting string \"%s\" into tokens:\n",str);
 pch = strtok (str," ,.-");
 while (pch != NULL)
 {
 printf ("%s\n",pch);
 pch = strtok (NULL, " ,.-");
 }
 return 0;
}
</code></pre>

It's not C++, but it will work

blocks|key|580596|text|实际上，这是一个糟糕的答案，但我怀疑你所说的手动拆分的10行循环。有4行代码可以帮我做到：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|580597|#include+<vector>
#include+<iostream>
int+main()+{
++++using+std::vector;

++++const+char+foo[]+=+"meh\0heh\0foo\0bar\0frob";

++++vector<vector<char>+>+strings(1);
++++for+(const+char+*it=foo,+*end=foo%2Bsizeof(foo);+it!=end;+%2B%2Bit)+{
++++++++strings.back().push_back(*it);
++++++++if+(*it+==+'\0')+strings.push_back(vector<char>());
++++}

++++std::cout+<<+"number+of+strings:+"+<<+strings.size()+<<+'\n';
++++for+(vector<vector<char>+>::iterator+it=strings.begin(),+end=strings.end();+
+++++++++it!=end;+%2B%2Bit)
++++++++std::cout+<<+it->data()+<<+'\n';
}|code-block|syntax|javascript|580598|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|$]]

A bad answer, actually, but I doubted your claim of a 10 line loop for manual splitting. 4 Lines do it for me:

<pre><code>#include &lt;vector&gt;
#include &lt;iostream&gt;
int main() {
 using std::vector;

 const char foo[] = "meh\0heh\0foo\0bar\0frob";

 vector&lt;vector&lt;char&gt; &gt; strings(1);
 for (const char *it=foo, *end=foo+sizeof(foo); it!=end; ++it) {
 strings.back().push_back(*it);
 if (*it == '\0') strings.push_back(vector&lt;char&gt;());
 }

 std::cout &lt;&lt; "number of strings: " &lt;&lt; strings.size() &lt;&lt; '\n';
 for (vector&lt;vector&lt;char&gt; &gt;::iterator it=strings.begin(), end=strings.end(); 
 it!=end; ++it)
 std::cout &lt;&lt; it-&gt;data() &lt;&lt; '\n';
}
</code></pre>

blocks|key|580663|text|下面是我自己提出的解决方案，假设缓冲区紧跟在最后一个字符串之后结束：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|580664|std::vector<std::string>+split(const+std::vector<char>&+buf)+{
++++auto+cur+=+buf.begin();
++++while+(cur+!=+buf.end())+{
++++++++auto+next+=+std::find(cur,+buf.end(),+'\0');
++++++++drives.push_back(std::string(cur,+next));
++++++++cur+=+next+%2B+1;
++++}
++++return+drives;
}|code-block|syntax|javascript|580665|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|$]]

Here's the solution I came up with myself, assuming the buffer ends immediately after the last string:

<pre><code>std::vector&lt;std::string&gt; split(const std::vector&lt;char&gt;&amp; buf) {
 auto cur = buf.begin();
 while (cur != buf.end()) {
 auto next = std::find(cur, buf.end(), '\0');
 drives.push_back(std::string(cur, next));
 cur = next + 1;
 }
 return drives;
}
</code></pre>

blocks|key|796750|text|一个更优雅和实际的解决方案(与我的其他答案相比)使用getline，并归结为2行只有C%2B%2B2003，并且不需要手动循环记账和调节：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|796751|#include+<iostream>
#include+<sstream>
#include+<string>

int+main()+{
++++const+char+foo[]+=+"meh\0heh\0foo\0bar\0frob";

++++std::istringstream+ss+(std::string(foo,+foo+%2B+sizeof+foo));
++++std::string+str;

++++while+(getline+(ss,+str,+'\0'))
++++++++std::cout+<<+str+<<+'\n';
}|code-block|syntax|javascript|796752|然而，请注意，基于范围的字符串构造函数已经指出了在-‘\0’处拆分的固有问题:您必须知道确切的大小，或者为终极终结符找到一些其他的字符组合。|796753|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

A more elegant and actual solution (compared to my other answer) uses getline and boils down to 2 lines with only C++2003, and no manual loop bookkeeping and conditioning is required:

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

int main() {
 const char foo[] = "meh\0heh\0foo\0bar\0frob";

 std::istringstream ss (std::string(foo, foo + sizeof foo));
 std::string str;

 while (getline (ss, str, '\0'))
 std::cout &lt;&lt; str &lt;&lt; '\n';
}
</code></pre>

However, note how the range based string constructor already indicates an inherent problem with splitting-at-'\0's: You must know the exact size, or find some other char-combo for the Ultimate Terminator.

blocks|key|796845|text|下面的代码依赖于具有隐式构造函数的std::string，该构造函数接受const+char*，从而使循环成为非常简单的两行代码：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|796846|#include+<iostream>
#include+<string>
#include+<vector>

template<+std::size_t+N+>
std::vector<std::string>+split_buffer(const+char+(&buf)[N])
{
++++std::vector<std::string>+result;

++++for(const+char*+p=buf;+p!=buf%2Bsizeof(buf);+p%2B=result.back().size()%2B1)
++++++++result.push_back(p);

++++return+result;
}

int+main()
{
++++std::vector<std::string>+test+=+split_buffer("wrgl\0brgl\0frgl\0srgl\0zrgl");

++++for+(auto+it+=+test.begin();+it+!=+test.end();+%2B%2Bit)
++++++++std::cout+<<+'"'+<<+*it+<<+"\"\n";

++++return+0;
}|code-block|syntax|javascript|796847|这个解决方案假设缓冲区的大小是已知的，并且字符串列表的结尾条件是已知的。如果列表由"\0\0"终止，则需要将循环中的条件从p!=foo%2Bsizeof(foo)更改为*p。|796848|entityMap^0|H|B|10|B|0|0|15|6|1P|I|2A|2|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@$9|V|A|W|B|C]|$9|X|A|Y|B|C]|$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|11|8|@]|D|@]|E|$]]]|N|$]]

The following relies on <code>std::string</code> having an implicit constructor taking a <code>const char*</code>, making the loop a very simple two-liner: 

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

template&lt; std::size_t N &gt;
std::vector&lt;std::string&gt; split_buffer(const char (&amp;buf)[N])
{
 std::vector&lt;std::string&gt; result;

 for(const char* p=buf; p!=buf+sizeof(buf); p+=result.back().size()+1)
 result.push_back(p);

 return result;
}

int main()
{
 std::vector&lt;std::string&gt; test = split_buffer("wrgl\0brgl\0frgl\0srgl\0zrgl");

 for (auto it = test.begin(); it != test.end(); ++it)
 std::cout &lt;&lt; '"' &lt;&lt; *it &lt;&lt; "\"\n";

 return 0;
}
</code></pre>

This solution assumes the buffer's size is known and the criterion for the end of the list of strings. If the list is terminated by <code>"\0\0"</code> instead, the condition in the loop needs to be changed from <code>p!=foo+sizeof(foo)</code> to <code>*p</code>.

I have a series of strings stored in a single array, separated by nulls (for example ['f', 'o', 'o', '\0', 'b', 'a', 'r', '\0'...]), and I need to split this into a <code>std::vector&lt;std::string&gt;</code> or similar.

I could just write a 10-line loop to do this using <code>std::find</code> or <code>strlen</code> (in fact I just did), but I'm wondering if there is a simpler/more elegant way to do it, for example some STL algorithm I've overlooked, which can be coaxed into doing this.

It is a fairly simple task, and it wouldn't surprise me if there's some clever STL trickery that can be applied to make it even simpler.

Any takers?

Simple way to split a sequence of null-separated strings in C++

我有一系列存储在单个数组中的字符串，由空值分隔(例如，'f'，'o'，'o'，'\0'，'b'，'a'，'r'，'\0'...)，我需要将其拆分为std::vector<std::string>或类似的值。我可以使用std::find或strlen编写一个10行的循环来完成这项工作(实际上我刚刚就做到了)，但我想知道是...

问在C++中拆分以空值分隔的字符串序列的简单方法
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