c++ - 如何在C++ 11或更高版本中使用带有类和重载<<操作符的枚举？

我正在学习C++，这是我的代码:

#include <iostream>
#include <vector>
using namespace std;

enum class Genre {
    unknown, fiction, nonfiction, periodical, biography, children
};

class Book {
    public:
        Genre getGenre() {return gen;}
        void setGenre(Genre gBook) {gen=gBook;}
private:
    Genre gen;
};

vector<Book> books;

void readInBooks() {        // loop a few times to get some elements in books
    int genre;
    cout << "Please enter genre (1 fiction, 2 nonfiction, 3 periodical, 4 biography, 5 children): ";
    cin >> genre;
    Book b;
    Genre g=Genre(genre);
    b.setGenre(g);
    books.push_back(b);
}

void printBooks() {
    cout << "Total library:" << endl << endl;
    for(int i=0;i<books.size();i++) {
        cout << "Book " << i << ": " << books[i].getGenre() << endl;
    }
}

int main() {
    readInBooks();
    printBooks();
    return 0;
}

所以这行:

 cout << "Book " << i << ": " << books[i].getGenre() << endl;

给出此错误:

操作符<
所以我尝试了:

 friend ostream& operator<<(ostream& os, Genre& g) {
    os << g.getGenre();
    return os;
 }

在Class Book的公开部分，出现错误:

g中属于非类类型Genre的成员getGenre的请求。

感谢您向我解释我做错了什么。

最佳答案

您的运算符与枚举有关-它与Book无关，并且不应与Book类有任何关系

因此，在定义枚举后，您可以定义用于打印它的运算符

enum class Genre {
    unknown, fiction, nonfiction, periodical, biography, children
};
std::ostream& operator<< (std::ostream& os, const Genre& g)
{
    switch (g) {
    case Genre::unknown:
        return os << "unknown";
    case Genre::fiction:
        return os << "fiction";
    case Genre::nonfiction:
        return os << "nonfiction";
    case Genre::periodical:
        return os << "periodical";
    case Genre::biography:
        return os << "biography";
    case Genre::children:
        return os << "children";
    }
    return os << static_cast<int>(g); //this line should be unreachable
    // - but for future extention of enum if you forget to extend the switch as well
    //I have added a protection
}