我有几个类似这样的类(class):
struct neg_inf {
constexpr double operator()() { return -std::numeric_limits<double>::infinity(); }
};
struct pos_inf {
constexpr double operator()() { return std::numeric_limits<double>::infinity(); }
};
template<typename dX, class LowerBound, class UpperBound>
class limit {
dX dx;
UpperBound upperBound;
LowerBound lowerBound;
double step_size;
limit( dX x, LowerBound lower, UpperBound upper, double step = 1 ) :
dx{ x }, lowerBound{ lower }, upperBound{ upper }, step_size{ step }
{}
dX value() const { return dx; }
LowerBound lower() const { return lowerBound; }
UpperBound upper() const { return upperBound; }
double step() const { return step_size; }
};
到目前为止,这些类都按预期工作。现在我想使用 std::enable_if_t
、std::is_arithemtic
std::is_same
等条件来修改模板参数。
这些是实例化限制对象需要满足的条件。
dX
must be at leastarithmetic
andnumerical
Lower
&Upper
bound
must be eithernumerical
,arithmetic
orneg_inf
orpos_inf
.
例如,这些是有效的实例化:
dX = 1st and can be any: int, long, float, double, /*complex*/, etc.
limit< 1st, 1st, 1st > // default template
limit< 1st, 1st, pos_inf > // these will be specializations
limit< 1st, 1st, neg_inf >
limit< 1st, pos_inf, 1st >
limit< 1st, neg_inf, 1st >
limit< 1st, pos_inf, pos_inf >
limit< 1st, neg_inf, neg_inf >
limit< 1st, neg_inf, pos_inf >
limit< 1st, pos_inf, neg_inf >
这些是实例化我的模板的有效条件。当 UpperBound
和/或 LowerBound
是 infinity
类型时,我计划部分专门化此类。当上限
和下限
是数值算术类型时,通用或默认模板将处理它们。
我的问题是,使用 type_traits
库,我的类的模板声明
会是什么样子?
最佳答案
将模板类型限制为类的一种方法是为 SFINAE 添加额外的参数:
template <typename dX, class LowerBound, class UpperBound, typename Enabler = void>
class limit;
然后,通过适当的 SFINAE 提供特化
template <typename dX, class LowerBound, class UpperBound>
class limit<dX,
LowerBound,
UpperBound,
std::enable_if_t<my_condition<dX, LowerBound, UpperBound>::value>>
{
// ...
};
因此,在您的情况下,my_condition
应该类似于
template <typename dX, class LowerBound, class UpperBound>
using my_condition =
std::conjunction<std::is_arithmetic<dX>,
std::disjunction<std::is_arithmetic<LowerBound>,
std::is_same<LowerBound, neg_inf>,
std::is_same<LowerBound, pos_inf>>,
std::disjunction<std::is_arithmetic<UpperBound>,
std::is_same<UpperBound, neg_inf>,
std::is_same<UpperBound, pos_inf>>
>;
另一种方式是static_assert
:
template <typename dX, class LowerBound, class UpperBound>
class limit
{
static_assert(std::is_arithmetic<dX>::value, "!");
static_assert(std::is_arithmetic<LowerBound>::value
|| std::is_same<LowerBound, neg_inf>::value
|| std::is_same<LowerBound, pos_inf>::value, "!");
static_assert(std::is_arithmetic<UpperBound>::value
|| std::is_same<UpperBound, neg_inf>::value
|| std::is_same<UpperBound, pos_inf>::value, "!");
// ...
};
关于c++ - 指定类模板参数的要求,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/56657847/