Scope resolution operator and const

Question

Let\'s take the following code:

#include  // std::string    
using namespace std;
int main() {
  //const::int i = 42;       -> Error: \"expected         


        

Answer 1

It compiles because

using namespace std;

pulls the entire std namespace into the global namespace, so ::string is the same as std::string. Your line in question is actually interpreted as

const ::string str = "Foo";

However, int is a keyword (and a fundamental type), and not a user-defined name that resides in any namespace. So ::int doesn't make sense.