当我偶然发现这个问题时,我正在尝试使用C++ 0x可变参数模板:
template < typename ...Args >
struct identities
{
typedef Args type; //compile error: "parameter packs not expanded with '...'
};
//The following code just shows an example of potential use, but has no relation
//with what I am actually trying to achieve.
template < typename T >
struct convert_in_tuple
{
typedef std::tuple< typename T::type... > type;
};
typedef convert_in_tuple< identities< int, float > >::type int_float_tuple;
当我尝试输入模板参数包时,GCC 4.5.0给出了一个错误.
基本上,我想将参数包"存储"在typedef中,而无需解压缩.可能吗？如果没有,是否有一些理由不允许这样做？
我目前正在为元组编写算术运算符重载.运算符迭代元组以对其每个元素执行操作.这是operator + =的定义:
template< typename... Ts, std::size_t I = 0 >
inline typename std::enable_if< I == sizeof... (Ts), std::tuple< Ts... >& >::type operator +=(std::tuple< Ts... >& lhs, const std::tuple< Ts... >& rhs)
{
return lhs;
}
template< typename... Ts, std::size_t I = 0 >
inline typename std::enable_if< I != sizeof... (Ts), std::tuple< Ts... >& >::type operator +=(std::tuple< Ts... >& lhs, const std::tuple< Ts... >& rhs)
{
std::get< I >(lhs) += std::get< I >(rhs);
return operator +=< Ts..., I + 1 …