为什么我的隐式转换函数不起作用？
我有一个包装类，我想修改数据并将其转换回其原始类型。
class A
{
public:
A ( unsigned __int64 _a ) : a (_a)
{
}
operator unsigned __int64 () const
{
return a;
}
unsigned __int64 a;
};
我希望此类的对象隐式转换回 unsigned __int64，但失败了。
比如说，
A a( 0x100ull );
unsigned __int64 b = (a >> 16); // Error
编译器给出 C2678 错误，找不到运算符或没有可接受的转换。 看来这个函数 operator unsigned __int64 () const 不起作用。
更具体地说，编译器说没有可接受的转换。我不能接受抱怨，因为我已经给出了很好的抱怨。有人可以使其合法化吗？
I have a wrapper class and I want to modify the data and convert it back to its original type.
class A
{
public:
A ( unsigned __int64 _a ) : a (_a)
{
}
operator unsigned __int64 () const
{
return a;
}
unsigned __int64 a;
};
I want the object of this class to implicitly convert back to unsigned __int64, but it failed.
Say,
A a( 0x100ull );
unsigned __int64 b = (a >> 16); // Error
Compiler gives C2678 error, no operator found or there is no acceptable conversion.
It seems this function operator unsigned __int64 () const doesn't work.
To be more specific, compiler says there is no acceptable conversion. I cannot accept the complain, because I have already given a good one. Can someone legitimize it?
如果你对这篇内容有疑问，欢迎到本站社区发帖提问 参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。
绑定邮箱获取回复消息
由于您还没有绑定你的真实邮箱，如果其他用户或者作者回复了您的评论，将不能在第一时间通知您！
发布评论
评论（1）
它不起作用，因为您还没有为您的类创建一个接受整数并对其执行某些操作的
operator>>重载。我猜您正在尝试对 int 进行右移，但我不确定重载您的
operator>>是一个好主意，因为这些运算符在上下文中像这样，通常用于流媒体。之后它可能会让代码的读者或维护者感到困惑。请参阅此处了解有关运算符重载的更多信息
也许重新考虑您的实施策略？
It doesn't work because you haven't created an
operator>>overload for your class that takes an integer and does something with it.I'm guessing you're trying to do a right shift on your int, but I'm not sure that overloading your
operator>>is a good idea for that, as these operators in a context like that, are normally used for streaming. It might confuse a reader or maintainer of your code afterwards.See here for more info on operator overloading
Perhaps rethink your implementation strategy?