例如,我有以下Java代码:
public class Main {
public static void main(String[] args) {
System.out.println(maker(Employee::new));
}
private static Employee maker(Supplier<Employee> fx) {
return fx.get();
}
}
class Employee {
@Override
public String toString() {
return "A EMPLOYEE";
}
}
C ++的等价物是什么?
答案 0 :(得分:4)
供应商是一个不带参数并返回某种类型的函数:您可以用std::function表示:
#include <iostream>
#include <functional>
#include <memory>
// the class Employee with a "print" operator
class Employee
{
friend std::ostream& operator<<(std::ostream& os, const Employee& e);
};
std::ostream& operator<<(std::ostream& os, const Employee& e)
{
os << "A EMPLOYEE";
return os;
}
// maker take the supplier as argument through std::function
Employee maker(std::function<Employee(void)> fx)
{
return fx();
}
// usage
int main()
{
std::cout << maker(
[]() { return Employee(); }
// I use a lambda here, I could have used function, functor, method...
);
return 0;
}
我没有在这里使用指针,也没有使用new来分配Employee:如果你想使用它,你应该考虑像std::unique_ptr这样的托管指针:
std::unique_ptr<Employee> maker(std::function<std::unique_ptr<Employee>(void)> fx)
{
return fx();
}
// ...
maker(
[]()
{
return std::make_unique<Employee>();
}
);
注意:对运营商的调用&lt;&lt;然后应该修改,因为maker将返回一个指针而不是一个对象。
答案 1 :(得分:1)
您可以使用C ++模板类来实现目标:
template<typename TR> class Supplier
{
private:
TR (*calcFuncPtr)();
public:
Supplier(TR(*F)())
{
calcFuncPtr = F;
}
TR get() const
{
return calcFuncPtr();
}
};
使用示例:
#include <string>
#include <iostream>
struct Employee
{
public:
std::string ToString()
{
return "A EMPLOYEE";
}
};
Employee maker(const Supplier<Employee>& fx)
{
return fx.get();
}
Employee GetInstanceFunc()
{
return Employee();
}
int _tmain(int argc, _TCHAR* argv[])
{
Employee employee(maker(GetInstanceFunc));
std::cout << employee.ToString();
return 0;
}
第maker(GetInstanceFunc)
行的隐式类型转换允许在没有显式模板实例化的情况下使用Supplier
类。