Boost.Any得到原始类型

Question

I need to cast an any variable to an original type. I need to do this:

int i = 10;
any a(i);
int& i2 = any_cast<int &>(a);

But I want that the type stores in any variable. And I write this:

int i = 10;
any a(i);
a::type_value& i2 = any_cast<a::type_value &>(a); // there is no actually type_value

How can I do something like that? Or how can I extract the original type from the any variable? Boost.variant is convenient either.

If I cannot do that, then I have another question what C++ techniques and libraries can store and get the type through a function to solve this issue?

Answer 1

C++ is a statically typed language. The type of a boost::any is a runtime value; any particular any could have any type. That's kinda the point.

There is no any::type_value , because that would have to be a compile time value. And any is a runtime construct.

Consider this:

void TakeAnAny(boost::any a)
{
  a::type_value& i2 = any_cast<a::type_value &>(a);
}

What type is any::type_value ? It is legal to call TakeAnAny with virtually any type. There is no single compile-time type that any::type_value could reduce to. And therefore, there is no way for the compiler to determine a type. Since C++ is statically typed, you're hosed.

The ultimate purpose of any is type-erasure. I have some value. And I want to pass that to some other function. This process will go through several different communication layers. But I don't necessarily want all of those different layers to know exactly what type I'm using. I only need myself and my intended destination to know the type. So you stick it in an any and you're fine. Everyone else just sees the any , and both of you know what it wraps.

This process only works because both the source and the destination know the real type of the value. If you don't know the type, then you shouldn't be using any . The purpose of any is not have a function sit there and cast it to a bunch of possible types (that's what boost::variant is for). The purpose is to erase the type from a function's signature.

This allows for things like generic messages and signals. You register some event handler with a system. You fire an event that takes an any as a parameter. The person firing the event knows that the "MouseClick" event always takes a vec2 as its parameter. So every "MouseClick" handler casts it to a vec2 . The "KeyPress" event would maybe pass a int32_t . So those handlers cast it to that type. And so forth. Everyone knows what type it actually takes.

This used to be done with void* . The problem there is that you have ownership issues ( any is a value, while void* is a pointer). Also, a void* is so type-erased that there's no way to check to see if your cast is correct. any is really just a type&value-safe void* ; it prevents you from casting to the wrong type.

You don't really want any . Your use case doesn't seem to want variant either. What you seem to want is a template . That's a different kind of thing, and it would let you do what you really want: have a function that can use any particular type, while still being able to know exactly what that type is.

Of course, templates have their own limitations.

Answer 2

您可以使用std::type_info的映射到std::function<void(boost::any const&)>对象来处理您知道并且想要处理的类型：您将在地图中找到该条目使用a.type()并调用相应的函数，该函数将知道如何处理参数。