我的问题是:
我有两个C++模块,即A和B,作为有活力连接的图书馆。 提供基本的数学功能和习俗例外类型。 B是使用A的较高层次模块。
B: 某些功能指A的功能,并试图追捕习俗例外A:MyExceptionFromA,以便将其改为B类:MyExceptionFromB(因为模块B的使用者不需要知道A的实施细节)。
只要我仍然留在C++领域,所有东西都会被罚款。 然而,如果我暴露了B:一些功能(通过增强的 p),则该例外情形在C++模块中不再出现。
我可以赶不上:滞后者,A:MyException 从A中得出,叫作(e)来检索正确的人名,因此,我知道,出现了正确的例外。 因此,我怀疑,问题在于把这一人为象征变成正确的例外。
我发现,该链接,其中解释说,“python利用[星号]模式开放推广单元,因此,推广模块作者不必知道其他推广单元可能使用什么符号”。 我怀疑这是问题/解决办法的一部分,但我并不知道有象征意义的解决办法来说明如何解决我的问题。
任何想法?
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