c++ - 如何在 C++ 中打印通用 map

Question

我正在尝试编写一个函数来打印通用 map 。这是我到目前为止所拥有的:

    template<typename map_key, typename map_val>
    void log(DEBUG_LEVEL level, std::map<map_key, map_val> _map) {
        if (level >= d_level) {
            for (std::map<map_key, map_val>::iterator it = _map.begin();
                    it != _map.end(); ++it)
                std::cout << it->first << " => " << it->second << '\n';
        }
    }

它不编译。

error: expected ';' before 'it'
error: 'it' was not declared in this scope

1. 为什么不能编译？
2. 有没有更好的写法？我找到了一个单行打印 vector here (第二个答案)我想知道是否有相同的 map ？

更新:我修复了 (1) 但仍在等待 (2) 的答案我喜欢 Galik 的回答，但是当我尝试将其添加为函数时出现错误:error:'std::ostream& Logger::operator<<(std::ostream&, const std::pair<const _Key, _Tp>&)' must take exactly one argument

这是我的 Logger 类:

extern DEBUG_LEVEL d_level;

class Logger {

public:

    Logger(const char * app_name);
    Logger(DEBUG_LEVEL level, char * app_name);
    void log(DEBUG_LEVEL level, const char* str, ...);

    template<typename vector_type>
    void log(DEBUG_LEVEL level, const std::vector<vector_type>& _vec,
            const std::string seperator = ", ") {
        if (level >= d_level) {
            std::cout << get_prefix_msg() << " ";
            change_color(level);
            std::cout << "[";
            std::copy(_vec.begin(), _vec.end(),
                    std::ostream_iterator<vector_type>(std::cout,
                            seperator.c_str()));
            std::cout << "]\n";
            printf(ANSI_COLOR_RESET);
        }
    }

    template<typename map_key, typename map_val>
    void log(DEBUG_LEVEL level, const std::map<map_key, map_val>& _map,
            const std::string seperator = ", ") {
        if (level >= d_level) {
            std::cout << get_prefix_msg() << " \n";
            change_color(level);
            std::cout << "[";
            for (typename std::map<map_key, map_val>::const_iterator it =
                    _map.begin(); it != _map.end(); ++it)
                std::cout << it->first << " => " << it->second << seperator;
            std::cout << "]\n";
            printf(ANSI_COLOR_RESET);
        }
    }

    /**
     * will log but without the application _app_name and without the debug level (but will have the right color)
     */
    void strip_log(DEBUG_LEVEL level, const char* str, ...);

    void change_verbosity(DEBUG_LEVEL level);

    static int msg_id;
private:
    const char * _app_name;
    void change_color(DEBUG_LEVEL level, bool is_strip = false);
    int get_next_msg_id();
    std::string get_prefix_msg();
    static DEBUG_LEVEL d_level;

};

Answer 1

typename 丢失。应该是

template<typename map_key, typename map_val>
void log(DEBUG_LEVEL level, const std::map<map_key, map_val>& _map) {
    if (level >= d_level) {
        for (typename std::map<map_key, map_val>::const_iterator it = _map.begin(); it != _map.end(); ++it)
            std::cout << it->first << " => " << it->second << '\n';
    }
}

或者在 C++11 中

template<typename map_key, typename map_val>
void log(DEBUG_LEVEL level, const std::map<map_key, map_val>& _map) {
    if (level >= d_level) {
        for (const auto& p : _map)
            std::cout << p.first << " => " << p.second << '\n';
    }
}