7

我想在我的程序中实现一个类似命令行的界面。所以我收到了遵循正常命令行语法的字符串(例如“-G foo -dp bar --help”)。由于我不想再次实现解析器,所以我想使用 Boost。

问题是:如何将字符串传递给 Boost 程序选项,而不是 argCount 和 argValues 的组合。我是否需要先将文本转换为数字 (argCount) 和 char* 数组 (argValues) 才能做到这一点?如果是的话......有没有一种简单的方法可以做到这一点?

提前致谢。

4

2 回答 2

8

一种方法是将标记std::string化为std::vector<std::string>,然后将结果传递给 Boost.ProgramOption 的command_line_parser。Boost.ProgramOption 的文档简要介绍了这种方法。此外,我在此答案的一部分中使用了类似的方法。

这是一个最小的完整示例:

#include <algorithm>
#include <iostream>
#include <iterator>
#include <string>
#include <vector>

#include <boost/bind.hpp>
#include <boost/program_options.hpp>
#include <boost/tokenizer.hpp>

// copy_if was left out of the C++03 standard, so mimic the C++11
// behavior to support all predicate types.  The alternative is to
// use remove_copy_if, but it only works for adaptable functors.
template <typename InputIterator,
          typename OutputIterator, 
          typename Predicate>
OutputIterator 
copy_if(InputIterator first,
        InputIterator last,
        OutputIterator result,
        Predicate pred)
{
  while(first != last)
  {
    if(pred(*first))
      *result++ = *first;
    ++first;
  }
  return result;
}

/// @brief Tokenize a string.  The tokens will be separated by each non-quoted
///        space or equal character.  Empty tokens are removed.
///
/// @param input The string to tokenize.
///
/// @return Vector of tokens.
std::vector<std::string> tokenize(const std::string& input)
{
  typedef boost::escaped_list_separator<char> separator_type;
  separator_type separator("\\",    // The escape characters.
                           "= ",    // The separator characters.
                           "\"\'"); // The quote characters.

  // Tokenize the intput.
  boost::tokenizer<separator_type> tokens(input, separator);

  // Copy non-empty tokens from the tokenizer into the result.
  std::vector<std::string> result;
  copy_if(tokens.begin(), tokens.end(), std::back_inserter(result), 
          !boost::bind(&std::string::empty, _1));
  return result;
}

int main()
{
  // Variables that will store parsed values.
  std::string address;
  unsigned int port;      

  // Setup options.
  namespace po = boost::program_options;
  po::options_description desc("Options");
  desc.add_options()
    ("address", po::value<std::string>(&address))
    ("port",    po::value<unsigned int>(&port))
    ;

  // Mock up input.
  std::string input = "--address 127.0.0.1 --port 12345";

  // Parse mocked up input.
  po::variables_map vm;
  po::store(po::command_line_parser(tokenize(input))
                .options(desc).run(), vm);
  po::notify(vm);

  // Output.
  std::cout << "address = " << address << "\n"
               "port = " << port << std::endl;
}

产生以下输出:

address = 127.0.0.1
port = 12345
于 2013-08-22T16:06:04.347 回答
0

boost::program_options有一个名为split_unix@FaceBro 的函数指出。它也适用于 Windows,因此以下是跨平台的,借用了已接受答案的主要示例结构:

int main()
{
  // Variables that will store parsed values.
  std::string address;
  unsigned int port;      

  // Setup options.
  namespace po = boost::program_options;
  po::options_description desc("Options");
  desc.add_options()
    ("address", po::value<std::string>(&address))
    ("port",    po::value<unsigned int>(&port))
    ;

  // Mock up input.
  std::string input = "--address 127.0.0.1 --port 12345";

  // Parse mocked up input.
  po::variables_map vm;
  po::store(po::command_line_parser(po::split_unix(input))
                .options(desc).run(), vm);
  po::notify(vm);

  // Output.
  std::cout << "address = " << address << "\n"
               "port = " << port << std::endl;
}
于 2021-01-07T20:13:47.783 回答