做网站运营工资多少seo是网络优化吗

前提认知：模板类继承模板类，是需要建立在假设的前提下的，如果没有这个”假设“，编译将会失败

1.书上举例

2.完整代码举例

#include <iostream>class MsgInfo {  };class BaseCompany
{
public:BaseCompany(){}~BaseCompany(){}
};class BaseCompany1
{
public:BaseCompany1(){}~BaseCompany1(){}
};template<typename Company>
class MsgSender 
{
public:void Clear(const MsgInfo& info) {}
};template<>
class MsgSender<BaseCompany1>{};template<typename Company>
class LoggingMsgSender : public MsgSender<Company> {
public:void SendClearMsg(const MsgInfo& info) {/* this->*/Clear(info);        }
};int main()
{LoggingMsgSender<BaseCompany> t;return 0;
}

• 编译器报错

main.cpp: In member function 'void LoggingMsgSender<Company>::SendClearMsg(const MsgInfo&)':
main.cpp:35:20: error: there are no arguments to 'Clear' that depend on a template parameter, so a declaration of 'Clear' must be available [-fpermissive]35 |         /* this->*/Clear(info);|                    ^~~~~
main.cpp:35:20: note: (if you use '-fpermissive', G++ will accept your code, but allowing the use of an undeclared name is deprecated)

3.三种解决方式，建立“假设”

• 使用this->clear(info)，告诉编译器假设父类有这个函数

template<typename Company>
class LoggingMsgSender : public MsgSender<Company> {
public:void SendClearMsg(const MsgInfo& info) {this->Clear(info);        }
};

• 使用using

template<typename Company>
class LoggingMsgSender : public MsgSender<Company> {using MsgSender<Company>::Clear;
public:void SendClearMsg(const MsgInfo& info) {Clear(info);        }
};

• 指定父类的函数

template<typename Company>
class LoggingMsgSender : public MsgSender<Company> {
public:void SendClearMsg(const MsgInfo& info) {MsgSender<Company>::Clear(info);        }
};