模板专业化(boost :: lexical_cast) [英] Template specialization (boost::lexical_cast)

问题描述

我想将lexical_cast方法扩展为vector<uint>类型，但是不起作用. 我尝试了以下代码:

I want to extend the lexical_cast method for vector<uint> types, but it's not working. I tried the following code:

#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;
    }
}

我遇到以下错误:

错误:"std :: string"的模板ID"lexical_cast" boost :: lexical_cast(std :: vector)’与任何内容都不匹配 模板声明

error: template-id ‘lexical_cast’ for ‘std::string boost::lexical_cast(std::vector)’ does not match any template declaration

推荐答案

可以通过重载operator<<来扩展词法转换.

Lexical cast can be extended by overloading operator<<.

问题是std::vector和uint是您的类型:它们是内置库或标准库.这使您无法在名称空间内进行重载或专门化.

Problem is that std::vector nor uint are your types: they're builtin or standard library. This makes it not-okay for you to overload or specialize inside the namespace.

真正的解决方案:

C ++支持强类型输入:

C++ is favors strong typing:

#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;
    }
};

奖励 lexical_cast现在可以正常运行了！

BONUS lexical_cast now magically works!

在Coliru上直播

#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机械手，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}; }

这些也可以与Boost Lexicalcast一起使用:

These also work with Boost Lexicalcast:

在Coliru上直播

#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

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