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模板特化(boost :: lexical_cast)

[英]Template specialization (boost::lexical_cast)

I want to extend the lexical_cast method for vector<uint> types, but it's not working. 我想扩展vector<uint>类型的lexical_cast方法,但它不起作用。 I tried the following code: 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;
    }
}

I got following error: 我收到以下错误:

error: template-id 'lexical_cast' for 'std::string boost::lexical_cast(std::vector)' does not match any template declaration 错误:'std :: string boost :: lexical_cast(std :: vector)'的template-id'lexical_cast'与任何模板声明都不匹配

Lexical cast can be extended by overloading operator<< . 可以通过重载operator<<来扩展词法转换。

Problem is that std::vector nor uint are your types: they're builtin or standard library. 问题是std::vectoruint是你的类型:它们是内置的或标准的库。 This makes it not-okay for you to overload or specialize inside the namespace. 这使您无法在命名空间内重载或专门化。

Real solution: 真正解决方案

Use a strong User-Defined Type 使用强大的用户定义类型

C++ is favors strong typing: C ++有利于强类型:

#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 now magically works! 奖励 lexical_cast现在神奇地工作!

Live On Coliru 住在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";

}

Prints 打印

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

Adapt for IO 适应IO

Use a custom IO Manipulator like eg How do I output a set used as key for a map? 使用自定义IO操纵器,例如如何输出用作地图键的集合?

#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}; }

These also work with Boost Lexicalcast: 这些也适用于Boost Lexicalcast:

Live On Coliru 住在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";
}

Prints 打印

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

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