访问的boost ::函数参数 [英] Access boost::function arguments

问题描述

是否有可能访问包含在一个boost ::函数类型的参数？

Is it possible to access the arguments contained in a boost::function type?

我希望能够获取函数的地址被调用，并提供该功能的参数值。

I'd like to be able to retrieve the address of the function to be called, and the values of the arguments provided for that function.

推荐答案

提振::函数擦除实现类型，但如果你知道它，你可以施放它;因为升压::功能是按值（==！=）的信息显然是可比较的。

boost::function erases the implementation type, but if you know it, you can cast to it; since boost::function are comparable by value (== !=) the information is clearly available.

它看起来像（从功能N的function_base超类），你可以得到实施的对象与

It looks like (from the function_base superclass of functionN) you can get the implementation object with:

f.target<concrete_functor_type>()

这将返回NULL，如果你提供了错误的具体类型。

Which will return NULL if you provided the wrong concrete type.

此外，在function_base（可能超出上述目标的方法没有帮助）：

Also in function_base (probably not helpful beyond the target method above):

public: // should be protected, but GCC 2.95.3 will fail to allow access
  detail::function::vtable_base* vtable;
  mutable detail::function::function_buffer functor;

虚函数表，您可以访问：

vtable gives you access to:

      struct vtable_base
      {
        void (*manager)(const function_buffer& in_buffer, 
                        function_buffer& out_buffer, 
                        functor_manager_operation_type op);
      };

这可以让你的函子的typeid的：

which can get you the typeid of the functor:

  case get_functor_type_tag:
    out_buffer.type.type = &typeid(F);
    out_buffer.type.const_qualified = in_buffer.obj_ref.is_const_qualified;
    out_buffer.type.volatile_qualified = in_buffer.obj_ref.is_volatile_qualified;
    return;
  }

function_buffer（仿）只可用于裁判到函数对象，约束（这是固定的）成员函数的师生比，和免费的功能，在这里你还没有绑定任何参数

function_buffer (functor) is only useful for refs to function objects, bound (this is fixed) member functions ptrs, and free functions, where you haven't bound any arguments

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