以下是生成用于添加 N 个向量的表达式模板的代码,但目前我只关注添加两个向量。
template < typename Operation, typename Va>
struct VectorExpression < Operation, Va >
{
friend Va;
typedef VectorExpression< Operation, Va> This;
const Operation& op;
const Va& vector;
inline VectorExpression ( Operation oper, const Va& first) : op(oper), vector(first) {};
inline auto size() const ->decltype( vector.size() ) {
return vector.size();
};
inline auto operator[]( size_t i ) const
->decltype( vector[i] ) {
return vector[i];
};
};
template < typename Operation, typename Va, typename ... Vs >
struct VectorExpression <Operation, Va, Vs...> : protected VectorExpression< Operation, Vs...>
{
friend Va;
typedef VectorExpression< Operation, Va, Vs...> This;
typedef VectorExpression< Operation, Vs... > Base;
const Operation op;
const Va& vector;
inline VectorExpression ( Operation oper, const Va& first, const Vs&... vs ) : VectorExpression< Operation, Vs...>(oper, vs...), op(oper), vector(first) {};
inline auto size() const ->decltype( vector.size() ) {
return vector.size();
};
inline auto operator[]( size_t i ) const //This line is where the error says it is happening
->decltype( op( vector[i], Base::operator[](i) ) ) {
return op( vector[i], Base::operator[](i) );
};
};
不幸的是,当我尝试编译前面的内容时,我收到以下错误消息:
error: cannot call member function 'decltype (((const VectorExpression<Operation, Va>*)this)->VectorExpression<Operation, Va>::vector[i])
VectorExpression<Operation, Va>::operator[](size_t) const
[with
Operation = adder<std::complex<double>, double>;
Va = Vector<double, false>;
decltype (((const VectorExpression<Operation, Va>*)this)->VectorExpression<Operation, Va>::vector[i]) = const double&; size_t = long unsigned int]' without object
我不确定问题出在哪里。这似乎是一个相当简单的递归定义operator[]
,将当前对象中的向量与所有基类向量相加。
有任何想法吗?我不相信其余的代码是相关的[编辑:但是可以在此处找到标头,并且可以在此处找到测试“main” ,但是如果您想查看它,请告诉我。此外,C++11 中是否有任何方法可以在位置打印出类型以调试此类问题?