Rom*_*eno 12 c++ templates typename
我希望派生类是模板类的后代.那个班级取决于后代的成员.总之,我希望这段代码能够编译:
struct IBootParams
{
virtual bool GetImmediate() = 0;
};
template <class T>
struct TBootBootParams
{
typename T::TransType transferType;
typename T::UseAbort_ useAbort;
bool GetImmediate()
{
if ( transferType == T::e1 )
{
return useAbort.someFlag;
}
return false;
}
};
struct BootBootParams : public TBootBootParams<BootBootParams>
{
enum SomeEnum
{
e1=0,
e2,
e3
};
struct UseAbort
{
bool someFlag;
char someMember;
int otherMember;
} useAbort;
typedef SomeEnum TransType;
typedef UseAbort UseAbort_;
};
struct BootAltBootParams : public TBootBootParams<BootAltBootParams>
{
enum SomeEnum
{
e1=5,
e2,
e3
};
struct UseAbort
{
bool someFlag;
long long someMember;
long long otherMember;
} useAbort;
typedef SomeEnum TransType;
typedef UseAbort UseAbort_;
};
int _tmain(int argc, _TCHAR* argv[])
{
BootBootParams bp;
BootAltBootParams bpa;
bool f = bp.GetImmediate();
f = bpa.GetImmediate();
}
Run Code Online (Sandbox Code Playgroud)
ybu*_*ill 10
你不能这样做.当编译器实例化时,TBootBootParams<BootBootParams>它尚未完全读取BootBootParams的整个定义,因此您无法从TBootBootParams的定义中访问其成员(TBootBootParams的成员函数是一个例外,因为它们稍后被实例化).
通常的解决方案是有一个特质类:
template<class T> struct TBootBootParams_traits;
template <class T>
struct TBootBootParams
{
typename TBootBootParams_traits<T>::TransType transferType;
typename TBootBootParams_traits<T>::UseAbort_ useAbort;
bool GetImmediate()
{
if ( transferType == TBootBootParams_traits<T>::e1 )
{
return useAbort.someFlag;
}
return false;
}
};
struct BootBootParams;
template<> struct TBootBootParams_traits<BootBootParams>
{
enum SomeEnum
{
e1=5,
e2,
e3
};
struct UseAbort
{
bool someFlag;
long long someMember;
long long otherMember;
};
typedef SomeEnum TransType;
typedef UseAbort UseAbort_;
};
struct BootBootParams :
public TBootBootParams<BootBootParams>,
public TBootBootParams_traits<BootBootParams>
{
UseAbort_ useAbort;
};
Run Code Online (Sandbox Code Playgroud)