Question

我有一个问题似乎已经在这里讨论过了： CPP templated member function specialization

但this->template的解决方案与我的示例无关。

以下代码失败：

error: invalid operands of types '<unresolved overloaded function type>' and 'int'
       to binary 'operator<'

使用gcc 4.8.1

class Base { public: virtual int Do(){return 0;} };
class State1: public Base {};
class State2: public Base {};

template <typename ... T> class SM;

template <class StateBase, class HeadState, class ... States >
class SM<StateBase, HeadState, States...> : public SM< StateBase, States...>
{
    protected:
        HeadState headState;
        template<int cnt> StateBase* GetNextState ( unsigned int index ) { return headState; }
};  

template <class StateBase, class HeadState>
class SM< StateBase, HeadState>
{
    protected:
        HeadState headState;
        template<int cnt> StateBase* GetNextState ( unsigned int index ) { return headState; }
};  

template <class StateBase, class ... States >
class TopSM: public SM< StateBase, States...>
{
    public:
        void DoIt()
        {
            // following code fails with 
            // error: invalid operands of types '<unresolved overloaded function type>' and 'int' to binary 'operator<'
            int nextState = this->template SM< StateBase, States...>::GetNextState <1>( 1 );
        }
};  

TopSM<Base, State1, State2> sm;

int main()
{
    sm.DoIt();
    return 0;
}

Answer 1

template之前需要另一个GetNextState。如果标识符和.，->或::之前有模板参数，并且它是依赖于模板参数的成员，则需要{{1 }}关键字消除小于号的歧义。

template

Answer 2

几乎在那里，你需要另一个template

int nextState = this->template SM< StateBase, States...>::template GetNextState <1>( 1 );
                                                          ~~~~~~~~

问题在于，因为GetNextState来自模板参数，所以它不知道它是静态变量，函数，模板函数还是其他任何东西。解析器需要继续，因此它假设它不是模板函数，因此<被解析为小于运算符，而不是模板参数列表的开头。从那里，解析器变得混乱，你得到关于无效操作数的错误>。

对模板化基类的模板化成员的调用失败

2 个答案: