[英]Overload operator in template struct
I have the following scenario: I have a struct template<typename CType, int D> struct Point
in which I want to overload the operators < and >. 我有以下场景:我有一个结构
template<typename CType, int D> struct Point
我要重载运算符<和>。 Here comes the catch and the point I am not sure about: I want different implementations of < and > depending on if CType is float/double or int. 接下来是我不确定的问题:我想要<和>的不同实现,具体取决于CType是float还是double还是int。 Right now I'm doing this using typid from typeinfo, but I feel like this is not elegant.
现在我正在使用typeinfo中的typid来做这件事,但我觉得这不优雅。 How do I go about doing this in a clean way?
我该如何以干净的方式做到这一点?
Here is one option (using non-member operators): 这是一个选项(使用非成员运算符):
template<typename CType, int D>
bool operator<( Point<CType, D> const &p1, Point<CType, D> const &p2)
{
// generic logic
}
template<int D> bool operator<( Point<float, D> const &p1, Point<float, D> const &p2 )
{
// logic for float
}
It might be possible to replace float
with an enable_if
to make a version that works for all types of a certain type trait (eg have a single specialization for all floating point types). 可以使用
enable_if
替换float
以生成适用于特定类型特征的所有类型的版本(例如,对所有浮点类型具有单一特化)。
#include <iostream>
#include <type_traits>
template <typename CType, int D>
struct Point
{
template <typename T = CType>
auto operator<(int t) -> typename std::enable_if<std::is_same<T, int>::value, bool>::type
{
std::cout << "int" << std::endl;
return true;
}
template <typename T = CType>
auto operator<(float t) -> typename std::enable_if<std::is_same<T, float>::value, bool>::type
{
std::cout << "float" << std::endl;
return true;
}
};
int main()
{
Point<int, 1> pi;
Point<float, 1> pf;
pi < 5;
pf < 3.14f;
pi < 3.14f; // forced to apply operator<(int)
}
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