如何升级: : 函数和升级: : 绑定工作

我不喜欢在我的代码中散布魔术盒子... ... 这两个类到底是如何工作的,它们基本上允许任何函数被映射到一个函数对象,即使函数 < > 有一个完全不同的参数设置为一个 IM 传递给 boost::bind的参数

它甚至可以使用不同的调用约定(例如,成员方法是 VC 下的 __thiscall,但是对于那些需要与 C 兼容的函数,“普通”函数通常是 __cdecl__stdcall

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boost::function allows anything with an operator() with the right signature to be bound as the parameter, and the result of your bind can be called with a parameter int, so it can be bound to function<void(int)>.

This is how it works (this description applies alike for std::function):

boost::bind(&klass::member, instance, 0, _1) returns an object like this

struct unspecified_type
{
... some members ...
return_type operator()(int i) const { return instance->*&klass::member(0, i);
}

where the return_type and int are inferred from the signature of klass::member, and the function pointer and bound parameter are in fact stored in the object, but that's not important

Now, boost::function doesn't do any type checking: It will take any object and any signature you provide in its template parameter, and create an object that's callable according to your signature and calls the object. If that's impossible, it's a compile error.

boost::function is actually an object like this:

template <class Sig>
class function
{
function_impl<Sig>* f;
public:
return_type operator()(argument_type arg0) const { return (*f)(arg0); }
};

where the return_type and argument_type are extracted from Sig, and f is dynamically allocated on the heap. That's needed to allow completely unrelated objects with different sizes bind to boost::function.

function_impl is just an abstract class

template <class Sig>
class function_impl
{
public:
virtual return_type operator()(argument_type arg0) const=0;
};

The class that does all the work, is a concrete class derived from boost::function. There is one for each type of object you assign to boost::function

template <class Sig, class Object>
class function_impl_concrete : public function_impl<Sig>
{
Object o
public:
virtual return_type operator()(argument_type arg0) const=0 { return o(arg0); }
};

That means in your case, the assignment to boost function:

  1. instantiates a type function_impl_concrete<void(int), unspecified_type> (that's compile time, of course)
  2. creates a new object of that type on the heap
  3. assigns this object to the f member of boost::function

When you call the function object, it calls the virtual function of its implementation object, which will direct the call to your original function.

DISCLAIMER: Note that the names in this explanation are deliberately made up. Any resemblance to real persons or characters ... you know it. The purpose was to illustrate the principles.