在命名空间中使用 boost::program_options::validate 和自己的类

Question

如果magic_number 不在命名空间中，则以下改编示例编译，但如果它在命名空间中，则编译失败。

如果目标类在命名空间中，为什么找不到validate 函数？

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MSVC上产生的错误是：

converter_lexical.hpp(243): 错误 C2338: 目标类型既不是 std::istream'able 或 std::wistream'able。

我在使用 Clang here 时遇到同样的错误。

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测试；评论/取消评论#define USE_NAMESPACE:

// Copyright Vladimir Prus 2002-2004.
// Distributed under the Boost Software License, Version 1.0.
// (See accompanying file LICENSE_1_0.txt
// or copy at http://www.boost.org/LICENSE_1_0.txt)

// This example shows how a user-defined class can be parsed using
// specific mechanism -- not the iostream operations used by default.
//
// A new class 'magic_number' is defined and the 'validate' method is overloaded
// to validate the values of that class using Boost.Regex.
// To test, run
//   
//    regex -m 123-456
//    regex -m 123-4567
// 
// The first invocation should output:
//
//   The magic is "456"
//
// and the second invocation should issue an error message.

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

namespace po = boost::program_options;

//#define USE_NAMESPACE

#ifdef USE_NAMESPACE
namespace ns1{
#endif

/* Define a completely non-sensical class. */
struct magic_number {
public:
    magic_number(int n) : n(n) {}
    magic_number() = default;
    int n;
};
#ifdef USE_NAMESPACE
}
#endif

/* Overload the 'validate' function for the user-defined class.
   It makes sure that value is of form XXX-XXX 
   where X are digits and converts the second group to an integer.
   This has no practical meaning, meant only to show how
   regex can be used to validate values.
*/
#ifdef USE_NAMESPACE
void validate(boost::any& v, const std::vector<std::string>& values, 
    ns1::magic_number*, int)
#else
void validate(boost::any& v, const std::vector<std::string>& values, 
    magic_number*, int)
#endif
{
    static boost::regex r("\\d\\d\\d-(\\d\\d\\d)");


    // Make sure no previous assignment to 'a' was made.
    po::validators::check_first_occurrence(v);
    // Extract the first string from 'values'. If there is more than
    // one string, it's an error, and exception will be thrown.
    const std::string& s = po::validators::get_single_string(values);

    // Do regex match and convert the interesting part to 
    // int.
    boost::smatch match;
    if (boost::regex_match(s, match, r)) {
#ifdef USE_NAMESPACE
        v = boost::any(
            ns1::magic_number(boost::lexical_cast<int>(match[1])));
#else
        v = boost::any(
            magic_number(boost::lexical_cast<int>(match[1])));
#endif
    } else {
        throw po::validation_error(po::validation_error::invalid_option_value);
    }        
}


int main(int ac, char* av[])
{
    try {
        po::options_description desc("Allowed options");
        desc.add_options()
            ("help", "produce a help screen")
            ("version,v", "print the version number")
#ifdef USE_NAMESPACE
            ("magic,m", po::
                value<ns1::magic_number>(), "magic value (in NNN-NNN format)")
#else
            ("magic,m", po::
                value<magic_number>(), "magic value (in NNN-NNN format)")
#endif
            ;

        po::variables_map vm;
        store(parse_command_line(ac, av, desc), vm);

        if (vm.count("help")) {
            std::cout << "Usage: regex [options]\n";
            std::cout << desc;
            return 0;
        }
        if (vm.count("version")) {
            std::cout << "Version 1.\n";
            return 0;
        }
        if (vm.count("magic")) {
#ifdef USE_NAMESPACE
            std::cout << "The magic is \"" << vm["magic"].as<ns1::magic_number>().n << "\"\n";
#else
            std::cout << "The magic is \"" << vm["magic"].as<magic_number>().n << "\"\n";
#endif
        }
    }
    catch(std::exception& e)
    {
        std::cout << e.what() << "\n";
    }    
}

Answer 1

在ns1 命名空间中定义validate() 允许它编译（或将magic_number 移出它，如您所见）。在 Boost 调用 validate() 的任何地方，名称查找似乎都不会考虑您的全局重载。

编辑：我认为这是因为boost::program_options 中定义的模板版本validate() 被认为比全局重载匹配更好，但比与magic_number 在同一命名空间中定义的重载匹配更差。如果我对此有误，也许对 C++ 名称查找规则更了解的人可以发表评论。