类 Python C++ 装饰器 [英] Python-like C++ decorators

问题描述

有没有办法像python风格一样在C++中装饰函数或方法?

Are there ways to decorate functions or methods in C++ like in python style?

@decorator
def decorated(self, *args, **kwargs):
     pass

以宏为例:

DECORATE(decorator_method)
int decorated(int a, float b = 0)
{
    return 0;
}

或

DECORATOR_MACRO
void decorated(mytype& a, mytype2* b)
{
}

有可能吗?

推荐答案

std::function 为我提出的解决方案提供了大部分构建块.

std::function provides most of the building blocks for my proposed solution.

这是我提出的解决方案.

Here is my proposed solution.

#include <iostream>
#include <functional>

//-------------------------------
// BEGIN decorator implementation
//-------------------------------

template <class> struct Decorator;

template <class R, class... Args>
struct Decorator<R(Args ...)>
{
   Decorator(std::function<R(Args ...)> f) : f_(f) {}

   R operator()(Args ... args)
   {
      std::cout << "Calling the decorated function.
";
      return f_(args...);
   }
   std::function<R(Args ...)> f_;
};

template<class R, class... Args>
Decorator<R(Args...)> makeDecorator(R (*f)(Args ...))
{
   return Decorator<R(Args...)>(std::function<R(Args...)>(f));
}

//-------------------------------
// END decorator implementation
//-------------------------------

//-------------------------------
// Sample functions to decorate.
//-------------------------------

// Proposed solution doesn't work with default values.
// int decorated1(int a, float b = 0)
int decorated1(int a, float b)
{
   std::cout << "a = " << a << ", b = " << b << std::endl;
   return 0;
}

void decorated2(int a)
{
   std::cout << "a = " << a << std::endl;
}

int main()
{
   auto method1 = makeDecorator(decorated1);
   method1(10, 30.3);
   auto method2 = makeDecorator(decorated2);
   method2(10);
}

输出:

Calling the decorated function.
a = 10, b = 30.3
Calling the decorated function.
a = 10

附言

Decorator 提供了一个地方，您可以在其中添加函数调用之外的功能.如果你想简单地传递到 std::function，你可以使用:

Decorator provides a place where you can add functionality beyond making the function call. If you want a simple pass through to std::function, you can use:

template<class R, class... Args >
std::function<R(Args...)> makeDecorator(R (*f)(Args ...))
{
   return std::function<R(Args...)>(f);
}

这篇关于类 Python C++ 装饰器的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！