(C++)为什么boost作者在这里使用结构而不是类？

Question

在boost 属性树的文档中,有一个正确使用的示例,在此处或在包中给出 libs/property_tree/examples/debug_settings.cpp.

我想知道的是关于这struct debug_settings条线.为什么要把它作为结构而不是类？它甚至有两个成员函数,load(...)和save(...).我认为提升作者有充分的理由这样做,并且它与...效率有某种关系,即使结构和类在"技术上"相同？

从列出的版权年份,我可以猜测这可能是C++ 98,C++ 03或C++ 0x,因此使用结构而不是类的原因至少来自于前C++ 11观点.

// ----------------------------------------------------------------------------
// Copyright (C) 2002-2006 Marcin Kalicinski
//
// Distributed under the Boost Software License, Version 1.0. 
// (See accompanying file LICENSE_1_0.txt or copy at 
// http://www.boost.org/LICENSE_1_0.txt)
//
// For more information, see www.boost.org
// ----------------------------------------------------------------------------

//[debug_settings_includes
#include <boost/property_tree/ptree.hpp>
#include <boost/property_tree/xml_parser.hpp>
#include <boost/foreach.hpp>
#include <string>
#include <set>
#include <exception>
#include <iostream>
namespace pt = boost::property_tree;
//]
//[debug_settings_data
struct debug_settings
{
    std::string m_file;               // log filename
    int m_level;                      // debug level
    std::set<std::string> m_modules;  // modules where logging is enabled
    void load(const std::string &filename);
    void save(const std::string &filename);
};
//]
//[debug_settings_load
void debug_settings::load(const std::string &filename)
{
    // Create empty property tree object
    pt::ptree tree;

    // Parse the XML into the property tree.
    pt::read_xml(filename, tree);

    // Use the throwing version of get to find the debug filename.
    // If the path cannot be resolved, an exception is thrown.
    m_file = tree.get<std::string>("debug.filename");

    // Use the default-value version of get to find the debug level.
    // Note that the default value is used to deduce the target type.
    m_level = tree.get("debug.level", 0);

    // Use get_child to find the node containing the modules, and iterate over
    // its children. If the path cannot be resolved, get_child throws.
    // A C++11 for-range loop would also work.
    BOOST_FOREACH(pt::ptree::value_type &v, tree.get_child("debug.modules")) {
        // The data function is used to access the data stored in a node.
        m_modules.insert(v.second.data());
    }

}
//]
//[debug_settings_save
void debug_settings::save(const std::string &filename)
{
    // Create an empty property tree object.
    pt::ptree tree;

    // Put the simple values into the tree. The integer is automatically
    // converted to a string. Note that the "debug" node is automatically
    // created if it doesn't exist.
    tree.put("debug.filename", m_file);
    tree.put("debug.level", m_level);

    // Add all the modules. Unlike put, which overwrites existing nodes, add
    // adds a new node at the lowest level, so the "modules" node will have
    // multiple "module" children.
    BOOST_FOREACH(const std::string &name, m_modules)
        tree.add("debug.modules.module", name);

    // Write property tree to XML file
    pt::write_xml(filename, tree);
}
//]

int main()
{
    try
    {
        debug_settings ds;
        ds.load("debug_settings.xml");
        ds.save("debug_settings_out.xml");
        std::cout << "Success\n";
    }
    catch (std::exception &e)
    {
        std::cout << "Error: " << e.what() << "\n";
    }
    return 0;
}

我已经看过StackOverflow上的一些帖子,但我真正想要的只是在这个例子中.我已经通读了

• 我什么时候应该使用结构而不是类？,
• 我应该为我的Linq2Sql数据使用Struct而不是轻量级数据类吗？,
• 何时在c ++中使用struct over class,
• 什么时候应该在C++中使用类vs结构？,
• C++中的结构与类,以及
• 使用类和结构 .

我的想法:对我来说,这看起来不像"普通旧数据(POD)",因为它有一个成员函数,因为它封装了基于类的对象std::string和std::set.那个字符串可以改变所以我在质疑"不变性".它有超过1个数据类型,可能大于2个字节.它具有访问加载和保存功能,使其不仅仅是一个简单的结构.Boost是一个C++库,因此它不应该指望有人会将它用于C语言库.

Answer 1

该类不封装任何数据; 它只是将它们聚合在一起.这与便利功能一起,看起来是课堂的重点.

Answer 2

他们可以做到这一点:

class debug_settings
{
public:
    std::string m_file;               // log filename
    int m_level;                      // debug level
    std::set<std::string> m_modules;  // modules where logging is enabled
    void load(const std::string &filename);
    void save(const std::string &filename);
};

但他们认为如果一切都是公开的,他们就不需要上课.

正如已经说过的那样:

struct和class在C++中是等价的,唯一的区别是默认情况下struct属性是public而class属性是private.