Visual C++ Compiler allows dependent-name as a type without "typename"?

Question

Today one of my friends told me that the following code compiles well on his Visual Studio 2008:

#include <vector>
struct A
{
  static int const const_iterator = 100;
};
int i;
template <typename T>
void PrintAll(const T & obj)
{
  T::const_iterator *i;
}
int main()
{
  std::vector<int> v;
  A a;
  PrintAll(a);
  PrintAll(v);
  return 0;
}

I usually use g++, and it always refuse to pass the second PrintAll() call. As I know, for this problem, g++ is doing the standard way translating a template.

So, is my knowledge wrong, or is it a extension of VS2008?

Answer 1

This is not an extension at all.

VC++ never implemented the two phases interpretation properly:

1. At the point of definition, parse the template and determine all non-dependent names
2. At the point of instantiation, check that the template produces valid code

VC++ never implemented the first phase... it's inconvenient since it means not only that it accepts code that is non-compliant but also that it produces an altogether different code in some situations.

void foo(int) { std::cout << "int" << std::endl; }

template <class T> void tfoo() { foo(2.0); }

void foo(double) { std::cout << "double" << std::endl; }

int main(int argc, char* argv[])
{
  tfoo<Dummy>();
}

With this code:

• compliant compilers will print "int", because it was the only definition available at the point of definition of the template and the resolution of foo does not depend on T.
• VC++ will print "double", because it never bothered with phase 1

It might seem stupid as far as differences go, but if you think about the number of includes you have in a large program, there is a risk that someone will introduce an overload after your template code... and BAM :/