Question

我想为lexical_cast类型扩展vector<uint>方法，但它不起作用。我尝试了以下代码：

#include <boost/lexical_cast.hpp>

namespace boost
{
    template <>
    inline string lexical_cast <string>(vector<uint> source)
    {
        string tmp;
        for (size_t i = 0; i < source.size(); ++i)
            if (i < source.size() - 1)
                tmp += boost::lexical_cast<string>(source[i]) + "|";
            else
                tmp += boost::lexical_cast<string>(source[i]);
        return tmp;
    }
}

我收到了以下错误：

错误：'std :: string的template-id'lexical_cast' boost :: lexical_cast（std :: vector）'与any不匹配模板声明

Answer 1

可以通过重载operator<<来扩展词法转换。

问题是std::vector和uint是您的类型：它们是内置的或标准库。这使得你无法在命名空间内重载或专门化。

真正的解决方案：

使用强大的用户定义类型

C ++更倾向于强类型：

#include <vector>
#include <ostream>

struct Source {
    std::vector<uint> _data;

    friend std::ostream& operator<<(std::ostream& os, Source const& s) {
        bool first = true;
        for(auto i : s._data) {
            if (!first) os << "|";
            first = false;
            os << i;
        }
        return os;
    }
};

BONUS lexical_cast现在神奇地工作了！

<强> Live On Coliru

#include <boost/lexical_cast.hpp>
#include <iostream>
#include <iomanip> // for std::quoted

int main() {
    Source s { {1,2,3,4,5} };
    std::cout << "Source is " << s << "\n";

    std::string text = boost::lexical_cast<std::string>(s);

    std::cout << "Length of " << std::quoted(text) << " is " << text.length() << "\n";

}

打印

Source is 1|2|3|4|5
Length of "1|2|3|4|5" is 9

适应IO

使用自定义IO操纵器，例如： How do I output a set used as key for a map?

#include <ostream>

template <typename Container>
struct pipe_manip {
    Container const& _data;

    friend std::ostream& operator<<(std::ostream& os, pipe_manip const& manip) {
        bool first = true;
        for(auto& i : manip._data) {
            if (!first) os << "|";
            first = false;
            os << i;
        }
        return os;
    }
};

template <typename Container>
pipe_manip<Container> as_pipe(Container const& c) { return {c}; }

这些也适用于Boost Lexicalcast：

<强> Live On Coliru

#include <boost/lexical_cast.hpp>
#include <iostream>
#include <set>
#include <vector>

int main() {
    std::vector<uint> s { {1,2,3,4,5} };
    std::cout << "Source is " << as_pipe(s) << "\n";

    std::string text = boost::lexical_cast<std::string>(as_pipe(std::set<std::string>{"foo", "bar", "qux"}));
    std::cout << "Other containers work too: " << text << "\n";
}

打印

Source is 1|2|3|4|5
Other containers work too: bar|foo|qux

模板特化（boost :: lexical_cast）

1 个答案:

使用强大的用户定义类型

适应IO