如果在 C++ 中有多个继承模板,则非法调用非静态成员函数
Illegal Call of Nonstatic Member Function if Multiple Inheriting Templates in C++
我的主机有 2 个策略,每个策略都有一个 print
函数。
如果我调用 Policy::print()
,没有问题,但如果我调用 OtherPolicy::print()
,就会出现错误。
Error C2352 'OtherPolicy<T,Host<T,SubPolicy,OtherPolicy>>::printer': illegal call of non-static member function
with
[
T=uint32_t
] Scratch D:\tmp\ScratchCpp\Scratch\Scratch\HostPolicy.h 63
#include <iostream>
template<
typename T,
class Host
>
class Policy {
public:
virtual void printer()
{
std::cout << "base policy" << std::endl;
}
};
template<
typename T,
class Host
>
class SubPolicy : Policy<T, Host> {
public:
void printer() override
{
auto host = static_cast<Host&>(*this);
std::cout << "sub policy" << std::endl;
}
};
template<
typename T,
class Host
>
class OtherPolicy {
public:
void printer()
{
auto host = static_cast<Host&>(*this);
std::cout << "other policy" << std::endl;
}
};
template<
typename T,
template<typename, class> class A,
template<typename, class> class B
>
class Host :
public A<T, Host<T, A, B>>,
public B<T, Host<T, Policy, B>> {
public:
void printer()
{
std::cout << "host" << std::endl;
A<T, Host>::printer();
B<T, Host>::printer(); // comment out this line to compile successfully
}
};
int main(int argc, char **argv)
{
Host<uint32_t, SubPolicy, OtherPolicy> host;
host.printer();
}
有人能解释一下这是怎么回事以及如何正确地做到这一点吗?
Host
的第二个碱基 class 的类型是 B<T, Host<T, Policy, B>>
, 不是 B<T, Host>
。这会导致错误,因为 B<T, Host>
不是 Host
的基础 class。
修复方法是正确命名基的类型 class:
B<T, Host<T, Policy, B>>::printer();
我的主机有 2 个策略,每个策略都有一个 print
函数。
如果我调用 Policy::print()
,没有问题,但如果我调用 OtherPolicy::print()
,就会出现错误。
Error C2352 'OtherPolicy<T,Host<T,SubPolicy,OtherPolicy>>::printer': illegal call of non-static member function
with
[
T=uint32_t
] Scratch D:\tmp\ScratchCpp\Scratch\Scratch\HostPolicy.h 63
#include <iostream>
template<
typename T,
class Host
>
class Policy {
public:
virtual void printer()
{
std::cout << "base policy" << std::endl;
}
};
template<
typename T,
class Host
>
class SubPolicy : Policy<T, Host> {
public:
void printer() override
{
auto host = static_cast<Host&>(*this);
std::cout << "sub policy" << std::endl;
}
};
template<
typename T,
class Host
>
class OtherPolicy {
public:
void printer()
{
auto host = static_cast<Host&>(*this);
std::cout << "other policy" << std::endl;
}
};
template<
typename T,
template<typename, class> class A,
template<typename, class> class B
>
class Host :
public A<T, Host<T, A, B>>,
public B<T, Host<T, Policy, B>> {
public:
void printer()
{
std::cout << "host" << std::endl;
A<T, Host>::printer();
B<T, Host>::printer(); // comment out this line to compile successfully
}
};
int main(int argc, char **argv)
{
Host<uint32_t, SubPolicy, OtherPolicy> host;
host.printer();
}
有人能解释一下这是怎么回事以及如何正确地做到这一点吗?
Host
的第二个碱基 class 的类型是 B<T, Host<T, Policy, B>>
, 不是 B<T, Host>
。这会导致错误,因为 B<T, Host>
不是 Host
的基础 class。
修复方法是正确命名基的类型 class:
B<T, Host<T, Policy, B>>::printer();