在你说OVERKILL之前,我不在乎。
如何让Boost.program_options处理所需的cat
选项-
?
我有
// visible
po::options_description options("Options");
options.add_options()("-u", po::value<bool>(), "Write bytes from the input file to the standard output without delay as each is read.");
po::positional_options_description file_options;
file_options.add("file", -1);
po::variables_map vm;
po::store(po::command_line_parser(argc, argv).options(options).positional(file_options).run(), vm);
po::notify(vm);
bool immediate = false;
if(vm.count("-u"))
immediate = true;
if(vm.count("file"))
support::print(vm["file"].as<vector<string>>());
在我运行cat - - -
时抛出异常:
无法识别的选项' - '
我希望它将-
视为位置参数,我需要在完整文件列表中以正确的顺序使用它。我怎么能做到这一点?
更新
我有一半修复。我需要
po::options_description options("Options");
options.add_options()("-u", po::value<bool>(), "Write bytes from the input file to the standard output without delay as each is read.")
("file", po::value< vector<string> >(), "input file");
po::positional_options_description file_options;
file_options.add("file", -1);
问题是,当输出参数时,我似乎只得到三个-
中的两个:
if(vm.count("file"))
support::print(vm["file"].as<vector<string>>());
support :: print很好地处理向量和东西。
答案 0 :(得分:4)
您需要定义位置的命名程序选项,在您的情况下是file
#include <boost/foreach.hpp>
#include <boost/program_options.hpp>
#include <iostream>
#include <string>
#include <vector>
namespace po = boost::program_options;
int
main( int argc, char* argv[] )
{
std::vector<std::string> input;
po::options_description options("Options");
options.add_options()
("-u", po::value<bool>(), "Write bytes from the input file to the standard output without delay as each is read.")
("file", po::value(&input), "input")
;
po::positional_options_description file_options;
file_options.add("file", -1);
po::variables_map vm;
po::store(po::command_line_parser(argc, argv).options(options).positional(file_options).run(), vm);
po::notify(vm);
bool immediate = false;
if(vm.count("-u"))
immediate = true;
BOOST_FOREACH( const auto& i, input ) {
std::cout << "file: " << i << std::endl;
}
}
如果您不想点击
,这是一个coliru demo和输出$ g++ -std=c++11 -O2 -pthread main.cpp -lboost_program_options && ./a.out - - -
file: -
file: -
file: -
如果你只看到3个位置参数中的2个,那很可能是因为argv[0]
按照惯例是程序名,因此不被认为是参数解析。这可以在basic_command_line_parser
模板
37 template<class charT>
38 basic_command_line_parser<charT>::
39 basic_command_line_parser(int argc, const charT* const argv[])
40 : detail::cmdline(
41 // Explicit template arguments are required by gcc 3.3.1
42 // (at least mingw version), and do no harm on other compilers.
43 to_internal(detail::make_vector<charT, const charT* const*>(argv+1, argv+argc+!argc)))
44 {}