std::any 没有 RTTI，它是如何工作的?

std::any without RTTI, how does it work?(std::any 没有 RTTI，它是如何工作的?)

问题描述

如果我想使用 std::any 我可以在关闭 RTTI 的情况下使用它.以下示例也使用 -fno-rtti 和 gcc 按预期编译和运行.

If I want to use std::any I can use it with RTTI switched off. The following example compiles and runs as expected also with -fno-rtti with gcc.

int main()
{   
    std::any x;
    x=9.9;
    std::cout << std::any_cast<double>(x) << std::endl;
}

但是std::any 是如何存储类型信息的呢?如我所见，如果我使用错误"类型调用 std::any_cast，我会按预期得到 std::bad_any_cast 异常.

But how std::any stores the type information? As I see, if I call std::any_cast with the "wrong" type I got std::bad_any_cast exception as expected.

这是如何实现的，或者这可能只是 gcc 功能?

How is that realized or is this maybe only a gcc feature?

我发现 boost::any 也不需要 RTTI，但我也没有发现它是如何解决的.boost::any 需要 RTTI 吗?.

I found that boost::any did also not need RTTI, but I found also not how that is solved. Does boost::any need RTTI?.

深入研究 STL 标头本身并没有给我答案.这段代码对我来说几乎无法阅读.

Digging into the STL header itself gives me no answer. That code is nearly unreadable to me.

推荐答案

TL;DR; std::any 持有一个指向模板化静态成员函数的指针班级.此函数可以执行许多操作，并且特定于给定类型，因为函数的实际实例取决于类的模板参数.

TL;DR; std::any holds a pointer to a static member function of a templated class. This function can perform many operations and is specific to a given type since the actual instance of the function depends on the template arguments of the class.

libstdc++中std::any的实现没那么复杂，你可以看看:

The implementation of std::any in libstdc++ is not that complex, you can have a look at it:

https://github.com/gcc-mirror/gcc/blob/master/libstdc%2B%2B-v3/include/std/any

基本上，std::any 包含两件事:

Basically, std::any holds two things:

• 指向(动态)分配的存储的指针；
• 指向存储管理器函数"的指针:

void (*_M_manager)(_Op, const any*, _Arg*);

当你用 T 类型的对象构造或分配一个新的 std::any 时，_M_manager 指向一个特定于type T(实际上是T特有的类的静态成员函数):

When you construct or assign a new std::any with an object of type T, _M_manager points to a function specific to the type T (which is actually a static member function of class specific to T):

template <typename _ValueType, 
          typename _Tp = _Decay<_ValueType>,
          typename _Mgr = _Manager<_Tp>, // <-- Class specific to T.
          __any_constructible_t<_Tp, _ValueType&&> = true,
          enable_if_t<!__is_in_place_type<_Tp>::value, bool> = true>
any(_ValueType&& __value)
  : _M_manager(&_Mgr::_S_manage) { /* ... */ }

由于此函数特定于给定类型，因此您不需要 RTTI 来执行 std::any 所需的操作.

Since this function is specific to a given type, you don't need RTTI to perform the operations required by std::any.

此外，在 std::any_cast 中很容易检查您是否正在转换为正确的类型.下面是std::any_cast的gcc实现的核心:

Furthermore, it is easy to check that you are casting to the right type within std::any_cast. Here is the core of the gcc implementation of std::any_cast:

template<typename _Tp>
void* __any_caster(const any* __any) {
    if constexpr (is_copy_constructible_v<decay_t<_Tp>>) {
        if (__any->_M_manager == &any::_Manager<decay_t<_Tp>>::_S_manage) {
            any::_Arg __arg;
            __any->_M_manager(any::_Op_access, __any, &__arg);
            return __arg._M_obj;
        }
    }
    return nullptr;
}

您可以看到它只是您尝试转换的对象内部的存储函数 (_any->_M_manager) 和您要转换的类型的管理器函数之间的简单相等性检查到 (&any::_Manager>::_S_manage).

You can see that it is simply an equality check between the stored function inside the object you are trying to cast (_any->_M_manager) and the manager function of the type you want to cast to (&any::_Manager<decay_t<_Tp>>::_S_manage).

_Manager<_Tp> 类实际上是 _Manager_internal<_Tp> 或 _Manager_external<_Tp> 的别名，具体取决于 _Tp.此类还用于为 std::any 类分配/构造对象.

The class _Manager<_Tp> is actually an alias to either _Manager_internal<_Tp> or _Manager_external<_Tp> depending on _Tp. This class is also used for allocation / construction of object for the std::any class.

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