c++ - how to get enum value from enum type? -

the question in title may sound trivial, better explain code want do...

in c++11 can this:

#include <iostream>  namespace x {     enum class  foo { a,b }; }  template <typename t> void foo(t t) {      if (t == t::a) { std::cout << "a"; } }  int main() {     foo(x::foo::a); } 

the important point here template foo not need know in namespace enum declared. call foo(y::foo::b) (provided there enum class called foo in namespace y having members a , b).

now question is: how same plain old enums (and c++98 stuff)?

this works:

#include <iostream>  namespace x {     enum foo { a,b }; }  template <typename t> void foo(t t) {      if (t == x::a) { std::cout << "a"; } }  int main() {     foo(x::a); } 

but because foo knows in namespace enum declared. , wont work y::foo::b ! (in c++11 works if replace line if (t == t::a) ..., plain enum)

is there way working in c++98/03 without refering x in template explicitly?

for sake of completeness, in c++98, this

template <typename t> void foo(t t) {      if (t == t::a) { std::cout << "a"; } } 

results in

error: ‘a’ not member of ‘x::foo’ 

ps: not allowed change enum , template has live in different namespace enum.

pps: simple if (t == 0) work, thats avoid

to add nicol bolas' answer, can kind of hack way solution using adl:

namespace x {     enum foo { a,b };     bool isa(foo t)     {         return t == a;     } }  template <typename t> void foo(t t) {      if (isa(t)) { std::cout << "a\n"; }     else{std::cout << "not a\n";} } 

live demo