Is it possible to do template specializations with the same body for unrelated types?
If the bodies of the 2 specializations are the same (even though A and B are totally unrelated types), is there a way to remove code duplication?Code:template <typename T> void f(T &x) { ... default implementation } template<> void f(A &x) { ... specialized implementation for T=A } template<> void f(B &x) { ... specialized implementation for T=B }
you can make some behavioral-implementation for group of types denoted for example by "T1" and play with std::enable_if to make this implementation for everything that is T1 (or not T).
If you give us some actual case, we'll be able to give an actual example.
Ah! That makes sense.
I think I can work it out from type traits and enable_if. Just needed the hint.
Thanks!
O_oI think I can work it out from type traits and enable_if. Just needed the hint.
You probably didn't really want to specialize function templates in any event.
The moral of the code is that specializations don't overload functions, qualifiers participate in both levels of resolution, and specialized function templates are templates to the compiler but not the linker.
Soma
Code:#include <iostream> struct SA{SA(){}}; struct SB{SB(){}}; template < typename F > void Dump1 ( F & f ) { std::cout << "g" << '\n'; } template <> void Dump1 ( SA & f ) { std::cout << "a" << '\n'; } template <> void Dump1 ( SB & f ) { std::cout << "b" << '\n'; } template < typename F > void Dump2 ( F f ) { std::cout << "g" << '\n'; } template < typename F > void Dump2 ( F * f ) { std::cout << "g *" << '\n'; } template <> void Dump2 ( SA * f ) { std::cout << "a *" << '\n'; } template <> void Dump2 ( SB * f ) { std::cout << "b *" << '\n'; } template < typename F > void Dump3 ( F f ) { std::cout << "g" << '\n'; } template <> void Dump3 ( SA * f ) { std::cout << "a *" << '\n'; } template <> void Dump3 ( SB * f ) { std::cout << "b *" << '\n'; } template < typename F > void Dump3 ( F * f ) { std::cout << "g *" << '\n'; } int main() { SA sA1; const SA sA2; SB sB1; const SB sB2; Dump1(sA1); Dump1(sA2); Dump1(sB1); Dump1(sB2); Dump2(&sA1); Dump2(&sA2); Dump2(&sB1); Dump2(&sB2); Dump3(&sA1); Dump3(&sA2); Dump3(&sB1); Dump3(&sB2); return(0); }
“Salem Was Wrong!” -- Pedant Necromancer
“Four isn't random!” -- Gibbering Mouther
