我是 C++ 模板的新手,所以请耐心等待。
我想做的是通过使用模板函数在我的类中实现某种策略模式。我认为这将内联策略。
我的理解是这可以通过仿函数来实现,但我不想引入新的类,我只想在我的类中内联策略函数。
可以说我有一堂课Calculator
。
计算器.h
#ifndef CALCULATOR_H
#define CALCULATOR_H
class Calculator
{
public:
Calculator();
virtual ~Calculator();
typedef void (*Strategy)(int param1, int param2);
void add(int param1, int param2);
template<class T>
void doStrategy(T strategy, int param1, int param2);
protected:
private:
};
#endif
计算器.cpp
Calculator::Calculator()
{
//ctor
}
Calculator::~Calculator()
{
//dtor
}
void
Calculator::add(int param1, int param2)
{
std::cout << "Sum " << param1+param2 << std::endl;
}
template<class T>
void
Calculator::doStrategy(T strategy, int param1, int param2)
{
strategy(param1,param2);
}
主文件
int main()
{
Calculator calc = Calculator();
calc.doStrategy<Calulator::Strategy>(calc.add,2,3);
return 0;
}
这失败了
error: no matching function for call to ‘Calculator::doStrategy(<unresolved overloaded function type>, int, int)’|
note: candidate is:|
note: template<class T> void Calculator::doStrategy(T, int, int)|
note: template argument deduction/substitution failed:|
note: cannot convert ‘calc.Calculator::add’ (type ‘<unresolved overloaded function type>’) to type ‘void (*)(int, int)’|
==稍后编辑==
主文件
typedef void (Calculator::*Strategy)(int, int);
int main()
{
Calculator calc = Calculator();
Strategy strategy = &Calculator::add;
calc.doStrategy<Strategy>(strategy,2,3);
return 0;
}
仍然失败:
undefined reference to `void Calculator::doStrategy<void (Calculator::*)(int, int)>(void (Calculator::*)(int, int), int, int)'