为什么std :: any_cast不支持隐式转换？ [英] why doesn't std::any_cast support implicit conversion?

问题描述

为什么 std :: any_cast 从实际存储中进行隐式转换时会引发 std :: bad_any_cast 异常

Why does std::any_cast throw a std::bad_any_cast exception when an implicit conversion from the actual stored type to the requested type would be possible?

例如：

std::any a = 10;  // holds an int now
auto b = std::any_cast<long>(a);   // throws bad_any_cast exception

为什么不允许这样做，是否有解决方法允许隐式转换（如果 std :: any 拥有的确切类型未知）？

Why is this not allowed and is there a workaround to allow an implicit conversion (in case the exact type that std::any holds is unknown)?

推荐答案

std :: any_cast 是根据 typeid 指定的。在此引用 cppreference ：

std::any_cast is specified in terms of typeid. To quote cppreference on this:

如果请求的 typeid 抛出 std :: bad_any_cast ValueType 与
不匹配操作数的内容。

Throws std::bad_any_cast if the typeid of the requested ValueType does not match that of the contents of operand.

由于 typeid 不不允许实现弄清楚隐式转换是可能的，据我所知， any_cast 也无法知道这是可能的。

Since typeid doesn't allow the implementation to "figure out" an implicit conversion is possible, there's no way (to my knowledge) that any_cast can know it's possible either.

否则， std :: any 提供的类型擦除依赖于仅在运行时可用的信息。而且该信息并不像编译器用于确定转换的信息那样丰富。这就是C ++ 17中类型擦除的代价。

To put it otherwise, the type erasure provided by std::any relies on information available only at run-time. And that information is not quite as rich as the information the compiler has for figuring out conversions. That's the cost of type erasure in C++17.

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