Is it somehow possible to define an overloaded template function based on a non-type parameter?
following situation:
template<uint8_t> void SetupMem();
template<> void SetupMem<4>()
{ /* some code */ }
template<> void SetupMem<8>()
{ /* some code */ }
void TemplateCaller()
{
// use the plattform-specific template function at compile-time
SetupMem<sizeof(size_t)>();
}
now is it somehow possible to change the return value of SetupMem based on the non-type parameter? e.g.:
template<> uint32_t SetupMem<4>(){}
template<> uint64_t SetupMem<8>(){}
So that TemplateCaller() does not explicitly calls SetupMem with the desired template parameter (so avoiding something like: SetupMem<uint64, sizeof(size_t)>();)? Possible solutions upto C++11 are welcome :)
Just use simple function overloading and std::integral_constant:
std::uint32_t SetupMem(std::integral_constant<int, 4>); // (0)
std::uint64_t SetupMem(std::integral_constant<int, 8>); // (1)
void TemplateCaller()
{
auto a = SetupMem(std::integral_constant<int, 4>{}); // calls (0)
auto b = SetupMem(std::integral_constant<int, 8>{}); // calls (1)
}
You can introduce a template type alias for readability:
template <int X>
using ic = std::integral_constant<int, X>;
std::uint32_t SetupMem(ic<4>);
std::uint64_t SetupMem(ic<8>);
void TemplateCaller()
{
auto a = SetupMem(ic<4>{});
auto b = SetupMem(ic<8>{});
}
If your compiler doesn't support integral_constant, all you need to do is define it yourself:
template <int>
struct ic { };
std::uint32_t SetupMem(ic<4>);
std::uint64_t SetupMem(ic<8>);
void TemplateCaller()
{
auto a = SetupMem(ic<4>{});
auto b = SetupMem(ic<8>{});
}
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