什么时候可以返回不同的type_info实例为同一类型？

问题描述

在：

Andrei继续：

即使标准不能保证这一点，常见的编译器，如GCC和Visual Studio？

假设 typeid 调用），每个应用程序，每个转换单元，每dll / so，或完全不同的一个表？

有时候& typeid（T）！=& typeid（T）？

我主要感兴趣的是Windows编译器，

解决方案

是的。在Windows下，DLL不能有未解析的符号，因此。如果您有：

foo.h

  struct foo {virtual 〜foo（）{}}; 
  

dll.cpp

  #includefoo.h
 ... 
 foo f; 
 cout<< & typeid（& f）<< endl 
  

main.cpp

  #includefoo.h
 ... 
 foo f; 
 cout<< & typeid（& f）<< endl 
  

会给你不同的指针。因为在dll被加载之前typeid（foo）应该存在
在dll和主exe

更多的是，在Linux下，如果主可执行文件没有编译-rdynamic（或--export-dynamic）然后typeid将被解析为不同的符号在可执行文件和
在共享对象（通常不会发生在ELF平台下），因为一些优化完成时链接可执行文件 - 删除不必要的符号。

Andrei Alexandrescu writes in Modern C++ Design:

Andrei continues:

Even though the standard does not guarantee this, how is this implemented in common compilers, such as GCC and Visual Studio?

Assuming typeid does not leak (and return a new instance every call), is it one "table" per application, per translation unit, per dll/so, or something completely different?

Are there times when &typeid(T) != &typeid(T)?

I'm mainly interested in compilers for Windows, but any information for Linux and other platforms is also appreciated.

解决方案

Yes. Under windows DLL can't have unresolved symbols, thus. If you have:

foo.h

struct foo { virtual ~foo() {} };

dll.cpp

#include "foo.h"
...
foo f;
cout << &typeid(&f) << endl

main.cpp

#include "foo.h"
...
foo f;
cout << &typeid(&f) << endl

Would give you different pointers. Because before dll was loaded typeid(foo) should existin both dll and primary exe

More then that, under Linux, if main executable was not compiled with -rdynamic (or --export-dynamic) then typeid would be resolved to different symbols in executable andin shared object (which usually does not happen under ELF platforms) because of some optimizations done when linking executable -- removal of unnecessary symbols.

这篇关于什么时候可以返回不同的type_info实例为同一类型？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！