c++ - 默认类型的动态转换

Question

我想知道为什么以下内容不起作用

 float f = 12;
 int& g = dynamic_cast<int&>(f);

我得到了错误

cannot dynamic_cast 'f' (of type 'float') to type 'int&' (target is not pointer or reference to class)

 int& g = dynamic_cast<int&>(f);

我知道动态转换处理指针和引用，所以我的问题是引用是否只适用于类类型而不适用于默认类型？

Answer 1

根据 C++ 标准 (#5.2.7.1):

1 The result of the expression dynamic_cast(v) is the result of converting the expression v to type T. T shall be a pointer or reference to a complete class type, or “pointer to cv void.”