使用' - '作为选项结束标记与boost :: program_options

问题描述

指示命令行程序选项结束的传统方法是使用选项 c。

Running this with a -- in the argument list causes a required_option exception. (I get similar results if instead of making a po::option for each trailing arg, I pack them all into po::option::original_tokens in one po::option.)

推荐答案

我有同样的问题，但放弃了。

I had this same question, but gave up on it.

我相信这样做的方法是调用 program_options :: command_line_parser :: extra_style_parser（），传递一个函数，它通过引用获取一个字符串的向量并返回一个 option c $ c> style_parser typedef in ）。

I believe the way to do this is to call program_options::command_line_parser::extra_style_parser(), passing it a function that takes a vector of string by reference and returns a vector of options (see the style_parser typedef in cmdline.hpp).

您的函数将需要检测到第一个令牌是 - ，创建一个新的选项对象，将输入中的所有其余标记放入选项的值向量中，并清空输入向量。在 libs / program_options / src / cmdline.cpp >

Your function will need to detect that the first token is "--", create a new option object, place all the rest of the tokens in the input into the option's value vector and empty the input vector. See program_options::detail::cmdline::parse_long_option, etc., in libs/program_options/src/cmdline.cpp for something to start with.

您可能需要注册一个特定的选项值，以便您可以轻松找到这个特殊的选项

You'll probably need to register a specific option value to use so that you can easily find this special option object at the end of the parsing and extract the set of additional non-option parameters from it.

我希望我能给你一些代码但我从来没有真正做到这一点，我最终只是采取附加参数一行一个stdin。

I wish I could give you some code but I never got around to actually doing this, I ended up just taking the additional parameters one-per-line on stdin.

编辑：

我觉得很难指出你没有解决问题的方向，所以我回去了，让它工作。问题是，您的位置参数条目未设置为接受多个令牌，并且您未填充值。 program_options 代码期望两者或它不工作。

I felt bad about pointing you in a direction that didn't solve the problem, so I went back and got it working. The problem is that your positional argument entry wasn't set up to accept multiple tokens and you weren't filling in the value. The program_options code expects both or it doesn't work.

以下是适用于我的完整代码： / p>

Here's the complete code that works for me:

#include <boost/program_options.hpp>
#include <iostream>

namespace po = boost::program_options;
typedef std::vector<std::string> stringvec;

std::vector<po::option> end_of_opts_parser(stringvec& args) {
  std::vector<po::option> result;
  stringvec::const_iterator i(args.begin());
  if (i != args.end() && *i == "--") {
    for (++i; i != args.end(); ++i) {
      po::option opt;
      opt.string_key = "pargs";
      opt.value.push_back(*i);      //  <== here
      opt.original_tokens.push_back(*i);
      result.push_back(opt);
    }
    args.clear();
  }
  return result;
}

int main(int argc, char* argv[])
{
    po::options_description desc("Allowed Options");
    desc.add_options()
        ("help,h", "produce help message")
        ("pargs", po::value<stringvec>()->multitoken(), "positional arguments");
    //                          and here     ^^^^^^^^^^^^^

    po::command_line_parser clparser(argc, argv);
    clparser.extra_style_parser(end_of_opts_parser);

    po::variables_map vm;
    po::store(clparser.options(desc).run(), vm);
    po::notify(vm);

    bool const help = !vm["help"].empty();
    std::cout << "help = " << help << " - ";

    // in addition, you don't need to use a separate vector of strings:
    stringvec const& pargs = vm["pargs"].as<stringvec>();
    std::copy(pargs.begin(), pargs.end(),
        std::ostream_iterator<std::string>(std::cout, ","));

    return 0;
}

运行时使用 -h - foo bar baz 它输出 help = 1 - foo，bar，baz，。

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