C++ 模板专门化（boost:：词法转换）

我想为

vector

类型扩展

lexical\u cast

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;
    }
}

#包括
名称空间提升
{
模板
内联字符串词法转换（向量源）
{
串tmp；
对于（size_t i=0；i

我得到了以下错误：

错误：“std:：string”的模板id“lexical\u cast” boost:：lexical_cast（std:：vector）“”与任何 模板声明

---

词法强制转换可以通过重载

运算符或添加操纵器方法来扩展，您不能更改正在专门化的函数模板的原型，而是通过
常量&
获取其参数。如果将参数类型更改为
vector const&
，代码将编译。但是使用下面答案中描述的另一种方法，专门化函数模板通常不是一个好主意，请参阅
#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;
    }
};

#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";

}

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

#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}; }

#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";
}

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