Function specialization to access struct members with getter using enum - c++

Question

I have an enum and a struct
enum STORE_ENUM { A_DATA, B_DATA, C_DATA, D_DATA };
struct Store {
int a;
char b;
long c;
bool d;
}
and I want to access its members with a specialized get function that basically looks like this
T get(STORE_ENUM,store s);
and it returns the appropriate type and hopefully statically type checks.
is this possible in C++?

Answer 1

Yes it's possible. The boost PFR library allows something very similar to how you've envisaged it, e.g.:
auto& x = boost::pfr::get<B_DATA>(s);
See the tutorial here.

Answer 2

With C++17, you can do the following
#include <iostream>
enum struct STORE_ENUM { A_DATA, B_DATA, C_DATA, D_DATA };
struct Store {
int a;
char b;
long c;
bool d;
};
template<STORE_ENUM storeEnum>
auto get(Store const & s){
if constexpr (storeEnum == STORE_ENUM::A_DATA) return s.a;
if constexpr (storeEnum == STORE_ENUM::B_DATA) return s.b;
if constexpr (storeEnum == STORE_ENUM::C_DATA) return s.c;
if constexpr (storeEnum == STORE_ENUM::D_DATA) return s.d;
}
int main(){
auto store = Store{ 0, 'a', 4l, true};
std::cout << get<STORE_ENUM::A_DATA>( store) << "\n";
std::cout << get<STORE_ENUM::B_DATA>( store) << "\n";
std::cout << get<STORE_ENUM::C_DATA>( store) << "\n";
std::cout << get<STORE_ENUM::D_DATA>( store) << "\n";
}
See https://godbolt.org/z/1vffh3

Answer 3

In my solution, we don't need enums. It uses templates.
#include <iostream>
#include <type_traits>
struct Store
{
int a;
char b;
long c;
bool d;
Store() //Default constructor
{
a = 0;
b = 0;
c = 0;
d = false;
}
Store(int a, char b, long c, bool d) //Constructor. Custom values.
{
this->a = a;
this->b = b;
this->c = c;
this->d = d;
}
template <typename T = int,
typename = typename std::enable_if<std::is_same<T, int>::value ||
std::is_same<T, char>::value ||
std::is_same<T, long>::value ||
std::is_same<T, bool>::value,
void>::type>
T GetData()
{
if (std::is_same<T, char>::value)
{
return b;
}
if (std::is_same<T, long>::value)
{
return c;
}
if (std::is_same<T, bool>::value)
{
return d;
}
return a;
}
};
int main()
{
Store store { 63, '#', 65, true };
std::cout << store.GetData() << std::endl;
std::cout << store.GetData<>() << std::endl;
std::cout << store.GetData<int>() << std::endl;
std::cout << store.GetData<char>() << std::endl;
std::cout << store.GetData<long>() << std::endl;
std::cout << std::boolalpha << store.GetData<bool>() << std::endl;
}
Output
63
63
63
#
65
true
Compile
clang++ ./main.cpp -std=c++11 or g++ ./main.cpp -std=c++11
Check/run this code in https://repl.it/#JomaCorpFX/FunctionSpecialization#main.cpp

Answer 4

std::tuple basically does this, and your type is basically a tuple. So the easy and fast way is to just reuse std::tuple's machinery.
In c++14 it might look like this:
template<STORE_ENUM e>
auto get( Store s ){
return std::get<(unsigned)e>(std::make_tuple(s.a,s.b,s.c,s.d));
}
template<STORE_ENUM e, class T>
void set( Store& s, T t ){
std::get<(unsigned)e>(std::tie(s.a,s.b,s.c,s.d))=t;
}
This is c++14, but the missing bit for c++11 is easy:
template<STORE_ENUM e>
using store_type = typename std::tuple_element<(unsigned)e, std::tuple<int,char,long,bool>>::type;
template<STORE_ENUM e>
store_type<e> get( Store s ) {
return std::get<(unsigned)e>(std::make_tuple(s.a,s.b,s.c,s.d));
}
template<STORE_ENUM e>
void set( Store& s, store_type<e> v ){
std::get<(unsigned)e>(std::tie(s.a,s.b,s.c,s.d))=v;
}