kiz*_*zx2 11 c++ templates metaprogramming
我想实现Composite模式:
class Animal
{
public:
virtual void Run() = 0;
virtual void Eat(const std::string & food) = 0;
virtual ~Animal(){}
};
class Human : public Animal
{
public:
void Run(){ std::cout << "Hey Guys I'm Running!" << std::endl; }
void Eat(const std::string & food)
{
std::cout << "I am eating " << food << "; Yummy!" << std::endl;
}
};
class Horse : public Animal
{
public:
void Run(){ std::cout << "I am running real fast!" << std::endl; }
void Eat(const std::string & food)
{
std::cout << "Meah!! " << food << ", Meah!!" << std::endl;
}
};
class CompositeAnimal : public Animal
{
public:
void Run()
{
for(std::vector<Animal *>::iterator i = animals.begin();
i != animals.end(); ++i)
{
(*i)->Run();
}
}
// It's not DRY. yuck!
void Eat(const std::string & food)
{
for(std::vector<Animal *>::iterator i = animals.begin();
i != animals.end(); ++i)
{
(*i)->Eat(food);
}
}
void Add(Animal * animal)
{
animals.push_back(animal);
}
private:
std::vector<Animal *> animals;
};
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你看,由于我对复合模式的简单要求,我最终写了很多相同的重复代码迭代在同一个数组上.
#define COMPOSITE_ANIMAL_DELEGATE(_methodName, _paramArgs, _callArgs)\
void _methodName _paramArgs \
{ \
for(std::vector<Animal *>::iterator i = animals.begin(); \
i != animals.end(); ++i) \
{ \
(*i)->_methodName _callArgs; \
} \
}
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现在我可以像这样使用它:
class CompositeAnimal : public Animal
{
public:
// It "seems" DRY. Cool
COMPOSITE_ANIMAL_DELEGATE(Run, (), ())
COMPOSITE_ANIMAL_DELEGATE(Eat, (const std::string & food), (food))
void Add(Animal * animal)
{
animals.push_back(animal);
}
private:
std::vector<Animal *> animals
};
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有没有办法用C++元编程"清洁"?
std::for_each有人建议作为解决方案.我认为我们的问题是更一般的问题的具体情况,让我们考虑一下我们的新宏:
#define LOGGED_COMPOSITE_ANIMAL_DELEGATE(_methodName, _paramArgs, _callArgs)\
void _methodName _paramArgs \
{ \
log << "Iterating over " << animals.size() << " animals"; \
for(std::vector<Animal *>::iterator i = animals.begin(); \
i != animals.end(); ++i) \
{ \
(*i)->_methodName _callArgs; \
} \
log << "Done" \
}
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看起来这个不能被替换 for_each
看看GMan的优秀答案,C++的这一部分肯定是非平凡的.就个人而言,如果我们只是想减少样板代码的数量,我认为宏可能是适合这种特殊情况的工具.
GMan建议std::mem_fun并std::bind2nd返回仿函数.不幸的是,这个API不支持3个参数(我不相信像这样的东西被释放到STL中).
为了便于说明,以下是使用委托函数boost::bind:
void Run()
{
for_each(boost::bind(&Animal::Run, _1));
}
void Eat(const std::string & food)
{
for_each(boost::bind(&Animal::Eat, _1, food));
}
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GMa*_*ckG 10
我不确定我是否真的看到了这个问题.为什么不是这样的:
void Run()
{
std::for_each(animals.begin(), animals.end(),
std::mem_fun(&Animal::Run));
}
void Eat(const std::string & food)
{
std::for_each(animals.begin(), animals.end(),
std::bind2nd(std::mem_fun(&Animal::Eat), food));
}
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还不错.
如果你真的想摆脱(小)样板代码,添加:
template <typename Func>
void for_each(Func func)
{
std::for_each(animals.begin(), animals.end(), func);
}
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作为私人实用程序成员,然后使用:
void Run()
{
for_each(std::mem_fun(&Animal::Run));
}
void Eat(const std::string & food)
{
for_each(std::bind2nd(std::mem_fun(&Animal::Eat), food));
}
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更简洁一点.不需要元编程.
实际上,元编程最终会失败.您正在尝试生成以文本方式定义的函数.元编程无法生成文本,因此您不可避免地会在某处使用宏来生成文本.
在下一级,您将编写该函数,然后尝试取出样板代码.std::for_each做得很好.当然,作为已经证明,如果你发现这是太多的重复,只是因素说出来为好.
作为回应LoggedCompositeAnimal评论中的示例,您最好的选择是做出类似于:
class log_action
{
public:
// could also take the stream to output to
log_action(const std::string& pMessage) :
mMessage(pMessage),
mTime(std::clock())
{
std::cout << "Ready to call " << pMessage << std::endl;
}
~log_action(void)
{
const std::clock_t endTime = std::clock();
std::cout << "Done calling " << pMessage << std::endl;
std::cout << "Spent time: " << ((endTime - mTime) / CLOCKS_PER_SEC)
<< " seconds." << std::endl;
}
private:
std::string mMessage;
std::clock_t mTime;
};
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这主要是自动记录操作.然后:
class LoggedCompositeAnimal : public CompositeAnimal
{
public:
void Run()
{
log_action log(compose_message("Run"));
CompositeAnimal::Run();
}
void Eat(const std::string & food)
{
log_action log(compose_message("Eat"));
CompositeAnimal::Eat(food);
}
private:
const std::string compose_message(const std::string& pAction)
{
return pAction + " on " +
lexical_cast<std::string>(animals.size()) + " animals.";
}
};
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像那样.有关lexical_cast的信息.