C ++显示有矢量的地图

Question

std::map<std::string, std::vector<string>> data;

为了通过使用copy打印出来，我的std::ostream_iterator应该如何？

显然std::ostream_iterator<std::pair<std::string, std::vector<std::string>>> out_it(std::cout, "\n");没有成功。

我的operator<<超载是以下std::ostream& operator<<(std::ostream& out, const std::pair<std::string, std::vector<std::string>>& p)，它写出p.first和p.second并返回它。

Answer 1

所以这里有一个operator<<，它将打印出你地图中一对的内容：

std::ostream& operator<<(std::ostream& out, const std::pair<std::string, std::vector<std::string>>& p) {
    out << p.first << ": "; // prints the string from key
    for (const auto& i : p.second) // loops throught the whole vector that is asociated with that key
        out << i << ", ";
    return out;
}

所以在这个例子中使用它。如果您在地图中输入此内容：

std::map<std::string, std::vector<string>> data;
std::vector<std::string> vec = {"VAL1", "VAL2", "VAL3"};
data.insert(std::make_pair("KEY", vec));
auto it = data.find("KEY");
std::cout << *it;

这将是使用上面的运算符<<打印出来的内容：

KEY: VAL1, VAL2, VAL3, 

您也可以稍微更改格式，以便逗号不在最后一个值之后，但这只是一个整容问题。你的问题在于你想要打印矢量，而它没有std operator <<。因此，为了打印矢量，你必须手动循环它的内容，就像我的示例中的ranged for。

Answer 2

如果你在C ++中进行任何认真的编程，最终你需要一种通用的方法来打印出集合。

以下是一个基础：

#include <iostream>
#include <map>
#include <vector>
#include <string>


// introduce the concept of an object that emits values to an ostream
// by default it simply calls operator <<
template<class T> struct emitter
{
    using arg_type = T;

    emitter(const T& v) : v_(v) {}

    friend std::ostream& operator<<(std::ostream& os, const emitter& e) {
        return os << e.v_;
    }

    const T& v_;
};

// introduce the concept of an io manipulator called emit
template<class T> auto emit(const T& v) -> emitter<T>
{
    return emitter<std::decay_t<T>>(v);
}

// specialise the emitter for maps
template<class K, class V, class C, class A>
struct emitter<std::map<K, V, C, A>>
{
    using arg_type = std::map<K, V, C, A>;

    emitter(const arg_type& v) : v_(v) {}

    friend std::ostream& operator<<(std::ostream& os, const emitter& e) {
        const char* elem_sep = "\n\t";
        const char* end_sep = " ";
        os << "{";
        for (const auto& elem : e.v_)
        {
            os << elem_sep << emit(elem.first) << ": " << emit(elem.second);
            end_sep = "\n";
        }

        return os << end_sep << "}";
    }

    const arg_type& v_;
};

// specialise the emitter for vectors
template<class V, class A>
struct emitter<std::vector<V, A>>
{
    using arg_type = std::vector<V, A>;

    emitter(const arg_type& v) : v_(v) {}

    friend std::ostream& operator<<(std::ostream& os, const emitter& e) {
        const char* elem_sep = " ";
        const char* end_sep = " ";
        os << "[";
        for (const auto& elem : e.v_)
        {
            os << elem_sep << emit(elem);
            elem_sep = ", ";
        }

        return os << end_sep << "]";
    }

    const arg_type& v_;
};


int main() {
    // build test data
    std::map<std::string, std::vector<std::string>> data;

    data.emplace("a", std::vector<std::string>{ "now", "is", "the", "time" });
    data.emplace("b", std::vector<std::string>{ "for", "all", "good", "men" });
    data.emplace("c", std::vector<std::string>{ "to", "come", "to", "the" });
    data.emplace("d", std::vector<std::string>{ "aid", "of", "their", "party" });

    // request an emitter manipulator
    std::cout << emit(data) << std::endl;
}

预期产量：

{
    a: [ now, is, the, time ]
    b: [ for, all, good, men ]
    c: [ to, come, to, the ]
    d: [ aid, of, their, party ]
}

Answer 3

要自定义打印矢量，您必须自己编写一些代码。为了确保使用自定义运算符，我建议您使用std::stringstream将键值对转换为字符串，然后将其输入std::ostream_iterator<std::string>。

像this（赦免using namespace std;）的东西：

#include <iostream>
#include <vector>
#include <string>
#include <iterator>
#include <algorithm>
#include <sstream>
#include <map>

using namespace std;

int main() {
    map<string, vector<string>> m {
        {"a", {"1", "2"}},
        {"b", {"1", "2"}},
        {"b", {"1", "2"}},
    };

    transform(begin(m), end(m), ostream_iterator<string>(cout), [](auto& p){
        stringstream ss;
        ss << p.first << ", {";
        bool first = true;
        for (auto& s : p.second)
        {
          ss << (first ? "" : ", ") << s;
          first = false;
        }
        ss << "}\n";
        return ss.str();
    });

    return 0;
}

我实际上并没有写过qazxsw poi，但是你可以用你的替换来使lambda体缩短）;