C ++ namespace alias

Question

I am using C ++ 11, while I need some classes from the C ++ 17 library. When using boost, from which the classes were added, I want to do the following:

#if __cplusplus < CPP17 using std::any = boost::any; #endif 

Such an alias is not allowed. Additionally, the std namespace extension causes undefined behavior . I want my code to look the same with respect to the C ++ version. Is there a clear way?

Answer 1

A clear way is to add a customized name for it.

 #if __cplusplus < CPP17 using my_any = boost::any; #else using my_any = std::any; #endif // using my_any... 

Answer 2

It seems like you are trying to create a type alias in a namespace. The corect syntax for this is:

 namespace namespace_name { using any = boost::any; } 

However, the standard prohibits adding definitions (there are exceptions for specialized templates) to the std , so if you try to define std::any , the behavior of your program will be undefined.

The use of any namespace, including global, is OK, but not those reserved for an implementation that includes std and its subheadings.

Answer 3

I do not see a big problem with undefined behavior. You use #if to check in C ++ 17, and you know any there before. If you really want this, I say, go after it and put an alias in std if it is earlier than C ++ 17.

At the end of the day, the helper functions / classes / etc will most likely be placed in another namespace or with the __ prefix, as this is available for standard libraries. I do not think that the pre-C ++ 17 implementation exports any to std.

There is no other way. Just ignore the "undefined behavior" and go if that works. There is nothing magical about destroying your code; the worst that can happen is the failed implementation of std, and it does not compile when defining an alias. In my opinion, some people face undefined behavior problems.