Macro for template definitions

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Macro for template definitions



I have a problem with alias template, because I have a code which must be compatible with VS 2012, which does not support alias template.



Let's say I have an alias template like:


template<typename A, typename B> using foo = bar<A,B>;



Then it would be very convenient to be able to do something like:


#ifdef NO_ALIAS_TEMPLATE_AVAILABLE
#define foo<x,y> bar<x,y>
#else
template<typename A, typename B> using foo = bar<A,B>;
#endif



However the best I can do is


#define foo(x,y) bar<x,y>



And I don't want to replace all template specializations in all my code by round brackets, for code readability.



Is there any way to have a macro with delimiters <> for its argument? Or is there no simple solution to my problem? If not, how to implement a strict equivalence to alias template?


<>




2 Answers
2



No, the preprocessor cannot use <> to delimit macro arguments.


<>



You can emulate an alias template with a class template:


template <typename A, typename B>
struct foo
typedef bar<A, B> type;
;

// Usage: foo<A, B>::type
// Generic context: typename foo<A, B>::type



Demo



This would make its usage less nice, but that's unavoidable.



The following simpler solution works for me.


#ifdef NO_ALIAS_TEMPLATE_AVAILABLE
#define foo bar
#else
template<typename A, typename B> using foo = bar<A,B>;
#endif



Given


template <typename A, typename B> struct bar ;



the following lines work fine.


foo<int, double> f; // Ok.
bar<int, double> b; // Ok.





This is a terrible idea for common names of foo, but it's actually not that bad if you are careful with your macro names: #define MY_LIBRARY_FOO ::fully::qualified::bar and template <typename A, typename B> using MY_LIBRARY_FOO = bar<A, B>
– Justin
Aug 8 at 17:54


foo


#define MY_LIBRARY_FOO ::fully::qualified::bar


template <typename A, typename B> using MY_LIBRARY_FOO = bar<A, B>






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