> On 16 Oct 2018, at 18:21, Akim Demaille <[email protected]> wrote:
>
>> Le 16 oct. 2018 à 14:57, Hans Åberg <[email protected]> a écrit :
>>
>> The minimal example below compiles with clang++6, but not with g++8. One
>> would think that it should define a qualified name B::A, used as
>> mu::B::A a;
>> as in the other cases below. But I have it probably by legacy since when
>> Bison used YYSTYPE only, though.
>
> Concretely, it is my understanding that this is irrelevant
> to Bison, isn’t it?
One can use templates—in the example below, B is the parser, and A the semantic
type. That would be using C++ paradigms rather than writing the code directly.
--
#include <iostream>
#include <string>
namespace mu {
class A {};
template<typename T>
class B {
public:
typedef T A;
};
}
int main () {
mu::B<mu::A>::A a;
mu::A b;
return 0;
}
--
