[英]how to use boost::any_cast (c++ library) to cast to base types?
I am using boost::any to have polymorphic types, I need to be able to cast an object to its base type. 我使用boost :: any来获得多态类型,我需要能够将一个对象转换为它的基类型。
class A {
public:
int x;
virtual int foo()= 0;
};
class B : public A {
public:
int foo() {
return x + 1;
}
};
int main() {
B* bb = new B();
boost::any any = bb;
bb->x = 44;
A* aa = boost::any_cast<A*>(any);
}
The code of the main function throws the following error at runtime: main函数的代码在运行时抛出以下错误:
terminate called after throwing an instance of 'boost::exception_detail::clone_impl<boost::exception_detail::error_info_injector<boost::bad_any_cast> >'
what(): boost::bad_any_cast: failed conversion using boost::any_cast
Abort trap
If I change the static_cast in the boost::any_cast code for reinterpret_cast it seems to work. 如果我在boost :: any_cast代码中为reinterpret_cast更改static_cast,它似乎可行。 However I am not sure about the consequences of that.
但是我不确定这会带来什么后果。
Do you have any ideas? 你有什么想法?
Upcasts (towards pointer-to-base) don't require an explicit cast in C++. 向上(朝向指针)不需要在C ++中进行显式转换。
On the other hand, boost::any_cast
will only succeed when casting to the exact same type as was originally stored. 另一方面,
boost::any_cast
只有在转换为与最初存储的完全相同的类型时才会成功。 (IIRC it uses typeid to check that you are trying to access the correct type, and typeid comparisons are only true for exact matches.) (IIRC使用typeid来检查您是否正在尝试访问正确的类型,而typeid比较仅适用于完全匹配。)
Hence: 因此:
A* aa = boost::any_cast<B*>(any);
However, it is somewhat unclear why you should be using boost::any
for polymorphic types. 但是,有些人不清楚为什么你应该使用
boost::any
来实现多态类型。 In particular, it is not a smart pointer and will not delete the stored object. 特别是,它不是智能指针,不会删除存储的对象。 More common is to store pointers to polymorphic object in a smart pointer, eg
boost::shared_ptr<A>
更常见的是在智能指针中存储指向多态对象的指针,例如
boost::shared_ptr<A>
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