解引用运算符重载如何真正起作用？

-Kevin - 我的电子邮件地址有效，但会定期更改。 要联系我，请使用地址来自最近的帖子。I''m sorry, I have absolutely no idea what you are asking.-Kevin--My email address is valid, but changes periodically.To contact me please use the address from a recent posting. 我真的不知道你想要做什么。也许发布一大块代码，主要编译并告诉我们你想做什么？ I don''t really get what you''re trying to do. Maybe posting a chunk o code that mostly compiles and show us what you''d like to do ? T * X< T> :: operator - >（）; 提供一级转换。我不想要那个。像 Y< T> X< T> :: operator - >（）; T * Y< T> :: operator - >（）; 我需要Y< T> ;温度生成。严格来说，我认为没有 重载 - >应该是需要的。类型转换应该足够 如果我有一个来自X< T>的转换路径了。到T *。 Joe SeighT * X<T>::operator ->();gives one level of conversion. I don''t want that. Something likeY<T> X<T>::operator ->();T * Y<T>::operator ->();I need a Y<T> temp generated. Strictly speaking, I would think nooverloading of -> should be needed. Type conversion should be sufficientif I have a conversion path from X<T> to T*.Joe Seigh 这篇关于解引用运算符重载如何真正起作用？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！ 上岸，阿里云！