如何解决如何使用Boost库反序列化并获得会员价值
首先,我想告诉你
我以here为例。在这里,代码的编写者通过使用类save
使用了load
和bus_schedule
函数。我遇到了在这里称为new_schedule
的对象的帮助下获取所有成员变量的问题。我假设这个对象将帮助我获取其他类的所有成员(也许在这一点上我完全错了。)
除此之外,我已经看到,大多数成员变量都在private
处,这意味着我无法从main
函数访问它们。这就是为什么,我设置了一些getter函数来返回一些成员值。例如,从bus_schedule
类开始,如果我可以返回schedule
列表,则可以访问trip_info
结构值以及类型为bus_route
的指针。
返回我使用过的schedule
列表的函数是:
std::list<std::pair<trip_info,bus_route *> > return_schedule()
{
return schedule;
}
在main
函数中,我使用了:
bus_schedule new_schedule;
auto returning_schedule = new_schedule.return_schedule();
for(auto i:returning_schedule)
{
cout<<(i.first)<<" ~~~~~~~ "<<(i.second)<<endl;
/*with i.first I can access the struct info,i.second pointer gives me the info of
bus_route pointer value ater dereferencing*/
}
但是在这里,我仍然坚持从i.second指针获取单独的值。我在这里知道,bus_route
类有一个名为stops
的列表。 Here也返回了此内容。但是在我看来,每次进行这种反向计算都是很麻烦的。
我的问题是,在反序列化之后,是否有从boost
直接获得所有成员变量的结果?我将为所有班级设置getter函数,但害怕找到使用这些函数的途径。
解决方法
这个问题实际上与序列化是分开的。这里没有任何序列化的东西提炼到数据结构。我对代码进行了现代化处理,以使用C ++ 17功能
-
auto
并移动语义 -
default
个特殊成员 -
[[nodiscard]]
,override
虚拟关键字 - 显式构造函数和
emplace
- 带解构的范围循环
-
make_unique<>
而不是new
/delete
- 扩展的聚合初始化程序
总而言之,它使代码更短,更简单,更有效¹。另外,它突出显示了您想知道的一些内容,我将在下面显示。
#include <iostream>
#include <list>
#include <memory>
#include <string>
struct gps_position {
int degrees;
int minutes;
float seconds;
friend std::ostream& operator<<(std::ostream& os,const gps_position& gp) {
return os << ' ' << gp.degrees << "º" << gp.minutes << "'" << gp.seconds << '"';
}
};
/////////////////////////////////////////////////////////////
// One bus stop
struct bus_stop {
gps_position latitude,longitude;
[[nodiscard]] virtual std::string description() const = 0;
virtual ~bus_stop() = default;
friend std::ostream& operator<<(std::ostream& os,const bus_stop& bs) {
return os << bs.latitude << bs.longitude << ' ' << bs.description();
}
protected:
explicit bus_stop(gps_position _lat,gps_position _long)
: latitude(_lat),longitude(_long) {}
};
/////////////////////////////////////////////////////////////
// Several kinds of bus stops
struct bus_stop_corner : bus_stop {
std::string street1,street2;
[[nodiscard]] std::string description() const override {
return street1 + " and " + street2;
}
explicit bus_stop_corner(gps_position _lat,gps_position _long,std::string _s1,std::string _s2)
: bus_stop(_lat,_long),street1(std::move(_s1)),street2(std::move(_s2)) {}
};
struct bus_stop_destination : bus_stop {
std::string name;
[[nodiscard]] std::string description() const override { return name; }
bus_stop_destination(gps_position _lat,std::string _name)
: bus_stop(_lat,name(std::move(_name)) {}
};
struct bus_route {
using bus_stop_pointer = bus_stop*;
std::list<bus_stop_pointer> stops;
void append(bus_stop* _bs) { stops.insert(stops.end(),_bs); }
friend std::ostream& operator<<(std::ostream& os,const bus_route& br) {
for (auto& stop : br.stops) {
os << '\n' << std::hex << "0x" << stop << std::dec << ' ' << *stop;
}
return os;
}
};
/////////////////////////////////////////////////////////////
// a bus schedule is a collection of routes each with a starting time
struct bus_schedule {
struct trip_info { int hour,minute; std::string driver; };
void append(const std::string& _d,int _h,int _m,bus_route* _br) {
schedule.emplace(schedule.end(),trip_info{_h,_m,_d},_br);
}
private:
friend std::ostream& operator<<(std::ostream& os,const bus_schedule& bs) {
for (auto const& [k,v] : bs.schedule) { os << k << *v; }
return os;
}
friend std::ostream& operator<<(std::ostream& os,const bus_schedule::trip_info& ti) {
return os << '\n' << ti.hour << ':' << ti.minute << ' ' << ti.driver << ' ';
}
std::list<std::pair<trip_info,bus_route*>> schedule;
};
int main() {
// fill in the data
// make a few stops
auto bs0 = std::make_unique<bus_stop_corner>(
gps_position{ 34,135,52.560F },gps_position{ 134,22,78.30F },"24th Street","10th Avenue");
auto bs1 = std::make_unique<bus_stop_corner>(
gps_position{ 35,137,23.456F },gps_position{ 133,35,54.12F },"State street","Cathedral Vista Lane");
auto bs2 = std::make_unique<bus_stop_destination>(
gps_position{ 35,136,15.456F },32,15.300F },"White House");
auto bs3 = std::make_unique<bus_stop_destination>(
gps_position{ 35,134,48.789F },16.230F },"Lincoln Memorial");
// make the schedule
bus_schedule original_schedule;
bus_route route0;
bus_route route1;
{
// make a route
route0.append(bs0.get());
route0.append(bs1.get());
route0.append(bs2.get());
// add trips to schedule
original_schedule.append("bob",6,24,&route0);
original_schedule.append("bob",9,57,&route0);
original_schedule.append("alice",11,2,&route0);
}
{
// make aother routes
route1.append(bs3.get());
route1.append(bs2.get());
route1.append(bs1.get());
// add trips to schedule
original_schedule.append("ted",7,17,&route1);
original_schedule.append("ted",38,&route1);
original_schedule.append("alice",47,&route1);
}
// display the complete schedule
std::cout << "schedule" << original_schedule;
}
打印:
schedule
6:24 bob
34º135'52.56" 134º22'78.3" 24th Street and 10th Avenue
35º137'23.456" 133º35'54.12" State street and Cathedral Vista Lane
35º136'15.456" 133º32'15.3" White House
9:57 bob
34º135'52.56" 134º22'78.3" 24th Street and 10th Avenue
35º137'23.456" 133º35'54.12" State street and Cathedral Vista Lane
35º136'15.456" 133º32'15.3" White House
11:2 alice
34º135'52.56" 134º22'78.3" 24th Street and 10th Avenue
35º137'23.456" 133º35'54.12" State street and Cathedral Vista Lane
35º136'15.456" 133º32'15.3" White House
7:17 ted
35º134'48.789" 133º32'16.23" Lincoln Memorial
35º136'15.456" 133º32'15.3" White House
35º137'23.456" 133º35'54.12" State street and Cathedral Vista Lane
9:38 ted
35º134'48.789" 133º32'16.23" Lincoln Memorial
35º136'15.456" 133º32'15.3" White House
35º137'23.456" 133º35'54.12" State street and Cathedral Vista Lane
11:47 alice
35º134'48.789" 133º32'16.23" Lincoln Memorial
35º136'15.456" 133º32'15.3" White House
35º137'23.456" 133º35'54.12" State street and Cathedral Vista Lane
结论
要方便地使用map
数据结构,可以使用循环as the example has it:
std::list<
std::pair<bus_schedule::trip_info,bus_route*>>::const_iterator it;
for (it = bs.schedule.begin(); it != bs.schedule.end(); it++) {
os << it->first << *(it->second);
}
但是,在C ++ 17中,您编写的内容与以下内容完全相同:
for (auto const& [info,route] : bs.schedule) { os << info << *route; }
您会看到[k,v]
使用结构化绑定来提取值类型的“第一”和“第二”部分。如果编译器中没有C ++ 17功能,则始终可以手动执行以下操作:
for (auto const& pair : bs.schedule) {
trip_info const& info = pair.first;
bus_route const& route = *pair.second;
os << info << route;
}
四处逛逛private
成员
我认为该数据结构不是典型的或“良好”的惯用C ++。使用原始指针会引发内存管理错误(例如,反序列化路由的每个停站都会泄漏)。
在此“陷阱”中使用吸气剂/阻气剂的风险为putting lipstick on a pig²。
我会选择将您需要的内容公开。这是一个示例,该示例可从main中打印所有细节,而不会重载operator<<
运算符:
// display the complete schedule
std::cout << "schedule";
for (auto const& [info,route] : original_schedule.schedule) {
std::cout << '\n' << info.hour << ':' << info.minute << ' ' << info.driver << ' ';
for (auto& stop : route->stops)
std::cout << '\n' << stop->latitude << stop->longitude << ' ' << stop->description();
}
与以前一样打印。
完整列表
#include <iostream>
#include <list>
#include <memory>
#include <string>
struct gps_position {
int degrees;
int minutes;
float seconds;
friend std::ostream& operator<<(std::ostream& os,longitude;
[[nodiscard]] virtual std::string description() const = 0;
virtual ~bus_stop() = default;
explicit bus_stop(gps_position _lat,name(std::move(_name)) {}
};
struct bus_route {
std::list<bus_stop*> stops;
void append(bus_stop* _bs) { stops.push_back(_bs); }
};
/////////////////////////////////////////////////////////////
// a bus schedule is a collection of routes each with a starting time
struct bus_schedule {
struct trip_info { int hour,_br);
}
std::list<std::pair<trip_info,&route1);
}
// display the complete schedule
std::cout << "schedule";
for (auto const& [info,route] : original_schedule.schedule) {
std::cout << '\n' << info.hour << ':' << info.minute << ' ' << info.driver << ' ';
for (auto& stop : route->stops)
std::cout << '\n' << stop->latitude << stop->longitude << ' ' << stop->description();
}
}
请注意,即使removing all blank lines and comments,and using the same formatting,代码也比示例短100行代码。
¹注意,该代码显然没有客观地“高效”,因为它可以进行所有形式的动态分配和运行时多态性
²这种反模式对于具有传统OOP背景的人来说很常见,请参见pseudo/quasi-classes(PDF)
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