这个有关C ++语法以及特定功能的问题。
我有一个Employee
类,带有2个数组成员函数。一个数组存储Employee
拥有的所有空闲时间,另一个数组存储Employee
涵盖的班次。我想定义一个帮助程序类,使我可以在两个数组中的任何一个上使用for范围循环(for (:)
)。例如,如果我这样做:
for (auto& ts : employee_freetime_iterator{ employee })
它将遍历员工的空闲时间。如果我这样做:
for (auto& ts : employee_shift_iterator{ employee })
它将遍历所有班次。我有一个定义如下的类:
template <typename T,
typename = std::enable_if_t<std::is_same_v<Employee, std::remove_cv_t<T>>>
>
struct employee_shift_iterator {
employee_shift_iterator(T& e);
};
在声明中,T
将是Employee
或const Employee
,而第3个模板参数是SFINAE来强制执行此操作。现在,如果我使用此类,那么我将不得不复制并粘贴两次,一次用于employee_freetime_iterator
,一次用于employee_shift_iterator
。为了减少代码冗余,我选择这样做:
enum ScheduleType {
FREE,
SHIFT,
ST_TOTAL
};
template <typename T, ScheduleType ST,
typename = std::enable_if_t<std::is_same_v<Employee, std::remove_cv_t<T>>>
>
struct employee_iterator {
constexpr static ScheduleType mScheduleType = ST;
employee_iterator(T& e);
};
现在,我可以使用ScheduleType
选择不同的帮助器函数,这些函数可以使我在Employee
类上进行迭代。我想做的是像这样创建2个不同类型的别名(每个ScheduleType
一个):
template <typename T>
using employee_freetime_iterator = employee_iterator<T, FREE>;
template <typename T, typename SFINAE>
using employee_shift_iterator = employee_iterator<T, SHIFT>;
但是如何转发构造函数参数,以便自动推断T
模板参数?只是按原样编译会给我这个错误:
src/main.cpp:47:40: error: missing template arguments before ‘{’ token
auto test = employee_freetime_iterator{ em };
em是我之前在代码中创建的一名员工。我已经重构了代码,并删除了不必要的部分并将其粘贴到下面。
enum ScheduleType {
FREE,
SHIFT,
ST_TOTAL
};
// Forward declaration
template <typename T, ScheduleType ST, typename SFINAE>
class employee_iterator;
struct Employee {
std::vector<TimeSlot> mFreeTime;
std::vector<TimeSlot> mShifts;
using timeslot_iterator = typename std::vector<TimeSlot>::iterator;
using timeslot_const_iterator = typename std::vector<TimeSlot>::const_iterator;
timeslot_const_iterator begin(const std::vector<TimeSlot>& s) const;
timeslot_const_iterator end(const std::vector<TimeSlot>& s) const;
timeslot_iterator begin(std::vector<TimeSlot>& s);
timeslot_iterator end(std::vector<TimeSlot>& s);
};
// Helper class
template <typename T, ScheduleType ST,
typename = std::enable_if_t<std::is_same_v<Employee, std::remove_cv_t<T>>>
>
struct employee_iterator {
using iterator = std::conditional_t<std::is_const_v<T>, Employee::timeslot_const_iterator, Employee::timeslot_iterator>;
constexpr static ScheduleType mScheduleType = ST;
std::add_pointer_t<T> mEmployee;
employee_iterator() = delete;
employee_iterator(T& e);
employee_iterator(T* e);
};
// Helper class c'tors
template <typename T, ScheduleType ST, typename SFINAE>
employee_iterator<T, ST, SFINAE>::employee_iterator(T& e)
: mEmployee{ &e }
{ }
template <typename T, ScheduleType ST, typename SFINAE>
employee_iterator<T, ST, SFINAE>::employee_iterator(T* e)
: mEmployee{ e }
{ }
// begin and end functions for iteration over Employee
template <typename T, ScheduleType ST, typename SFINAE>
typename employee_iterator<T, ST, SFINAE>::iterator begin(employee_iterator<T, ST, SFINAE> it) {
if constexpr (ST == FREE)
return it.mEmployee->begin(it.mEmployee->mFreeTime);
else
return it.mEmployee->begin(it.mEmployee->mShifts);
}
template <typename T, ScheduleType ST, typename SFINAE>
typename employee_iterator<T, ST, SFINAE>::iterator end(employee_iterator<T, ST, SFINAE> it) {
if constexpr (ST == FREE)
return it.mEmployee->end(it.mEmployee->mFreeTime);
else
return it.mEmployee->end(it.mEmployee->mShifts);
}
/// Type alias
template <typename T>
using employee_freetime_iterator = employee_iterator<T, FREE>;
template <typename T>
using employee_shift_iterator = employee_iterator<T, SHIFT>;
编辑:我知道我写的代码很长且令人困惑,所以我创建了一些很短的东西来显示我的问题。我该如何使用它?
#include <utility>
template <typename T1, typename T2>
using my_pair = std::pair<T1, T2>;
int main() {
// I can do this:
// will be inferred as std::pair<double, int>
std::pair test1{ 1.0, 5 };
// However the compiler has issues with this:
my_pair test2{1.0, 3};
}
答案 0 :(得分:1)
如果需要模板参数推导,则需要引入函数:
namespace detail{
// ... employee_iterator, ScheduleType etc, ...
template <typename T>
using employee_freetime_iterator = employee_iterator<T, FREE>;
template <typename T>
using employee_shift_iterator = employee_iterator<T, SHIFT>;
}
template<class T>
detail::employee_freetime_iterator<T> employee_freetime_iterator(T& e) {
return {e};
}
template<class T>
detail::employee_shift_iterator<T> employee_shift_iterator(T& e) {
return {e};
}
不幸的是,模板类型别名不允许类模板参数的推导。如果要放弃显式模板参数,则必须使用功能模板。