std :: any_cast而无需原始对象的类型
是否可以使用std :: Any_cast而无需放入第一个模板参数（对象的类型任何覆盖）？我尝试使用any_cast＆lt; exptype（typeId（tocast）.name（））＆gt;，但它不起作用。
还尝试从一开始就存储对象类型，但这也无效，因为变量无法存储类型。
Is it possible to use std::any_cast without putting in the first template argument (the type of the object the any is covering)? I tried using any_cast<decltype(typeid(toCast).name())> but it didn't work.
Also tried to store the objects types from the beginning, but that also didn't work because variables can't store types.
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C ++的基本原理之一是所有对象的类型在编译时已知。这是一个绝对的规则，也没有例外。
所讨论的对象的类型是
std ::任何。仅当该类型在编译时也已知时，它才能转换为其他类型。您会注意到
std :: type_info :: name（）不是constexpr表达式。返回的字符串仅在运行时才知道。您正在尝试将某些东西投入到只有在运行时才知道的对象。 C ++不起作用。通常，每当发生这种情况时，几乎所有时间都将涉及在基类中调用虚拟方法。您可能需要重新设计您的类以使用继承和虚拟方法；使用其公共基类代替
std ::任何;然后调用其虚拟方法。在某些情况下，std ::变体也可能起作用。One of the fundamentals principles of C++ is that the types of all objects are known at compile time. This is an absolute rule, and there are no exceptions.
The type of the object in question is
std::any. It is convertible to some other type only if that type is also known at compile time.You will note that
std::type_info::name()is not aconstexprexpression. The returned string is known only at run time. You are attempting to cast something to an object whose type would only be known at run time. C++ does not work this way.In general, whenever such a situation occurs, nearly all the time the correct solution will involve virtual methods getting invoked on a base class. You will likely need to redesign your classes to use inheritance and virtual methods; using their common base class instead of
std::any; then invoking its virtual methods. In some situationsstd::variantmay also work, too.