I have the unfortunate task in 2024 of adding some code to a codebase that needs to be used in a C++03 setting – where unfortunately the original class myclass can’t be changed.
I’ve gotten stuck on something simple which is accessing enum values from an enum that is defined in a nested class.
#include <cstdio>
namespace ns
{
template <typename T>
class myclass {
public:
enum etype {
e_option1,
e_option2,
e_option3,
e_option4
};
};
}
int main() {
typedef typename ns::myclass<int> mc_t;
mc_t::etype et = mc_t::etype::e_option2;
switch(et) {
case mc_t::etype::e_option1 : printf("option1n"); break;
case mc_t::etype::e_option2 : printf("option2n"); break;
case mc_t::etype::e_option3 : printf("option3n"); break;
case mc_t::etype::e_option4 : printf("option4n"); break;
}
return 0;
}
https://godbolt.org/z/KeK7M54Wb
<source>: In function 'int main()':
<source>:20:27: error: 'ns::mc_t::etype' is not a class or namespace
20 | mc_t::etype et = mc_t::etype::e_option2;
| ^~~~~
<source>:23:18: error: 'ns::mc_t::etype' is not a class or namespace
23 | case mc_t::etype::e_option1 : printf("option1n"); break;
| ^~~~~
<source>:24:18: error: 'ns::mc_t::etype' is not a class or namespace
24 | case mc_t::etype::e_option2 : printf("option2n"); break;
| ^~~~~
<source>:25:18: error: 'ns::mc_t::etype' is not a class or namespace
25 | case mc_t::etype::e_option3 : printf("option3n"); break;
| ^~~~~
<source>:26:18: error: 'ns::mc_t::etype' is not a class or namespace
26 | case mc_t::etype::e_option4 : printf("option4n"); break;
| ^~~~~
Compiler returned: 1
3
You don’t need the extra ::etype when accessing the enumerators in your example of unscoped enumeration as they are already visible in the enclosing scope(which is the class scope here).
So the modified working code looks like:
int main()
{
typedef typename ns::myclass<int> mc_t;
//-----------------------v------------->no need to use `etype` here.
mc_t::etype et = mc_t::e_option2;
switch(et) {
//---------------v--------------------------------------->no need to use etype here
case mc_t::e_option1 : printf("option1n"); break;
case mc_t::e_option2 : printf("option2n"); break;
case mc_t::e_option3 : printf("option3n"); break;
case mc_t::e_option4 : printf("option4n"); break;
}
return 0;
}
Working demo
You can access the enum values using just the class name, without using the enum name. Hence, instead of mc_t::etype::e_option2, you can access the value as just mc_t::e_option2.
I would assume this is because its not an enum class, so the values are accessible directly from the parent scope.