我能够对相似类型的点一起执行操作，但不能对不同类型的点执行操作。我想我需要一些方法将point int的矢量坐标转换为vector double的矢量坐标。
#include <vector>
#include <iostream>
#include <algorithm>
#include <functional>
#if 1
#define log(x) std::cout << x << std::endl;
#else
#define log(x)
#endif
template<typename type>
std::vector<type> operator+(const std::vector<type> l, const std::vector<type> r){
std::vector<type> ans;
std::transform(l.begin(), l.end(), r.begin(), std::back_inserter(ans), std::plus<type>());
return ans;
};
template<typename type>
std::vector<type> operator*(const std::vector<type> l, const std::vector<type> r){
std::vector<type> ans;
std::transform(l.begin(), l.end(), r.begin(), std::back_inserter(ans), std::multiplies<type>());
return ans;
};
template<typename type>
std::vector<type> operator-(const std::vector<type> l, const std::vector<type> r){
std::vector<type> ans;
std::transform(l.begin(), l.end(), r.begin(), std::back_inserter(ans), std::minus<type>());
return ans;
};
template<typename type>
std::vector<type> operator/(const std::vector<type> l, const std::vector<type> r){
std::vector<type> ans;
std::transform(l.begin(), l.end(), r.begin(), std::back_inserter(ans), std::divides<type>());
return ans;
};
template<class type> struct point{
template<typename a = type, typename... b> point(a coordinate, b... coordinates){
this->coordinates = {coordinate, coordinates...};
};std::vector<type> coordinates;
template<typename a = type> point(std::vector<a> coordinates){
this->coordinates = coordinates;
};
friend point<type> operator+(const point<type>& l, const point<type>& r){
point<type> ans(l.coordinates + r.coordinates);
return ans;
};
friend point<type> operator*(const point<type>& l, const point<type>& r){
point<type> ans(l.coordinates * r.coordinates);
return ans;
};
friend point<type> operator-(const point<type>& l, const point<type>& r){
point<type> ans(l.coordinates - r.coordinates);
return ans;
};
friend point<type> operator/(const point<type>& l, const point<type>& r){
point<type> ans(l.coordinates / r.coordinates);
return ans;
};
friend std::ostream& operator<<(std::ostream& stream, const point& p){
switch(p.coordinates.size()){
case 1:
std::cout << "(" << p.coordinates[0] << ")";
break;
default:
std::cout << "(";
for(int i = 0; i < p.coordinates.size(); ++i){
if(i == (p.coordinates.size() - 1))
std::cout << p.coordinates[i];
else
std::cout << p.coordinates[i] << ", ";
}
std::cout << ")";
break;
}
}
};
int main(){
point<int> a(2,5,5,4,3);
point<int> b(3,5,3,5,7);
point<double> c(a/b);
log(c);
return 0;
}
答案 0 :(得分：0)
您需要编写接受适当右侧的operator=的实现，或者您可以使用模板化的＆lt;＆gt;运算符=。这是C ++ 11的实现，如果您没有C ++ 11，那么您将需要第2版
template<typename T>
struct Point {
using value_type = T;
using self_type = Point<T>;
value_type m_value;
Point(value_type value) : m_value(value) {}
template<typename RhsT> self_type& operator = (const Point<RhsT>& rhs) {
m_value = static_cast<value_type>(rhs.m_value);
return *this;
}
};
int main() {
Point<int> a1(4);
Point<int> a2(5);
Point<double> a3(6.0);
a1 = a2;
a2 = a3;
return 0;
}
http://ideone.com/csATfH Pre-C ++ 11版本：
#include <string>
#include <type_traits>
template<typename T>
struct Point {
typedef T value_type;
typedef Point<T> self_type;
value_type m_value;
Point(value_type value) : m_value(value) {}
template<typename RhsT>
self_type& operator = (const Point<RhsT>& rhs) {
m_value = static_cast<value_type>(rhs.m_value);
return *this;
}
};
int main() {
Point<int> a1(4);
Point<int> a2(5);
Point<double> a3(6.0);
a1 = a2;
a2 = a3;
return 0;
}
带有enable_if的花哨版本：
#include <string>
#include <type_traits>
template<typename T>
struct Point {
using value_type = T;
using self_type = Point<T>;
value_type m_value;
Point(value_type&& value) : m_value(std::forward<value_type>(value)) {}
template<typename RhsT,
typename std::enable_if<
std::is_arithmetic<value_type>::value
&& std::is_arithmetic<RhsT>::value, int>::type = 0>
self_type& operator = (Point<RhsT>&& rhs) {
m_value = static_cast<value_type>(rhs.m_value);
return *this;
}
};
int main() {
Point<int> a1(4);
Point<int> a2(5);
Point<double> a3(6.0);
Point<std::string> a4("hello");
a1 = a2;
a2 = a3;
a3 = a4;
return 0;
}