逗号分隔令牌到const char **的向量 [英] Vector of comma separated token to const char**

问题描述

我正在尝试将逗号分隔的字符串转换为const char *的向量.使用以下代码，按预期输出是

I am trying to convert a comma separated string to vector of const char*. With the following code, by expected output is

ABC_
DEF
HIJ

但我明白了

HIJ
DEF
HIJ

我要去哪里错了?

代码:

#include <iostream>
#include <boost/tokenizer.hpp>
#include <vector>
#include <string>
using namespace std;

int main()
{
   string s("ABC_,DEF,HIJ");
   typedef boost::char_separator<char> char_separator;
   typedef boost::tokenizer<char_separator> tokenizer;

   char_separator comma(",");
   tokenizer token(s, comma);
   tokenizer::iterator it;

   vector<const char*> cStrings;

   for(it = token.begin(); it != token.end(); it++)
   {
      //cout << (*it).c_str() << endl;
      cStrings.push_back((*it).c_str());
   }

   std::vector<const char*>::iterator iv;
   for(iv = cStrings.begin(); iv != cStrings.end(); iv++)
   {
      cout << *iv << endl;
   }
   return 0;
}

http://ideone.com/3tvnUs

在以下答案的帮助下的解决方案: (PaulMcKenzie使用列表提供了一种更整洁的解决方案.)

Solution with help of answers below: (PaulMcKenzie offers a neater solution using lists.)

#include <iostream>
#include <boost/tokenizer.hpp>
#include <vector>
#include <string>
using namespace std;

char* createCopy(std::string s, std::size_t bufferSize)
{
   char* value = new char[bufferSize];
   memcpy(value, s.c_str(), (bufferSize - 1));
   value[bufferSize - 1] = 0;
   return value;
}

int main()
{
   string s("ABC_,DEF,HIJ");
   typedef boost::char_separator<char> char_separator;
   typedef boost::tokenizer<char_separator> tokenizer;

   char_separator comma(",");
   tokenizer token(s, comma);
   tokenizer::iterator it;

   vector<const char*> cStrings;

   for(it = token.begin(); it != token.end(); it++)
   {
      //cout << it->c_str() << endl;
      cStrings.push_back(createCopy(it->c_str(),
                                      (it->length() + 1)));
   }

   std::vector<const char*>::iterator iv;
   for(iv = cStrings.begin(); iv != cStrings.end(); iv++)
   {
      cout << *iv << endl;
   }

   //delete allocations by new
   //...
   return 0;
}

推荐答案

这就是问题:boost::tokenizer::iterator不会返回字符串副本的所有权，而是对内部副本的引用.

Here's the thing: boost::tokenizer::iterator doesn't return you ownership of a copy of the string, but a refernce to an internal copy.

例如，运行代码后，我得到:

For example, after running your code I get:

HIJ
HIJ
HIJ

解决方案是将cStrings.push_back((*it).c_str())替换为以下其中一项:

the solution is to replace cStrings.push_back((*it).c_str()) with one of the following:

    char* c = new char[it->length() + 1];
    c[it->length()] = 0;
    cStrings.push_back(c);
    std::strncpy(c, it->c_str(), it->length());

看起来不漂亮，但是您可能不会比这更快(至少如果您想使用boost::tokenizer.

doesn't look pretty, but you probably won't get faster than that (at least if you want to use boost::tokenizer.

另一种选择是将boost::tokenizer完全替换为例如strtok-可以在此处找到示例: C拆分char数组放入不同的变量

other option is to totally replace boost::tokenizer with e.g. strtok - an example can be found here: C split a char array into different variables

您也可以使用boost::algorithm::string::split，但是稍后可能需要将string重新映射到const char*.

you can also use boost::algorithm::string::split, but then you might need to remap string to const char* later on.

这篇关于逗号分隔令牌到const char **的向量的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！