Question

I was playing with Visual Studio and templates.

Consider this code

struct Foo
{
  struct Bar
  {
  };

  static const int Bar=42;
};

template<typename T>
void MyFunction()
{
  typename T::Bar f;  
}

int main()
{
    MyFunction<Foo>();
    return 0;
}

When I compile this is either Visual Studio 2008 and 11, I get the following error

error C2146: syntax error : missing ';' before identifier 'f'

Is Visual Studio correct in this regard ? Is the code violating any standards ?

If I change the code to

struct Foo
{
  struct Bar
  {
  };

  static const int Bar=42;
};


void SecondFunction( const int& )
{
}

template<typename T>
void MyFunction()
{
  SecondFunction( T::Bar ); 
}

int main()
{
    MyFunction<Foo>();
    return 0;
}

it compiles without any warnings. In Foo::BLAH a member preferred over a type in case of conflicts ?

EDIT Tests on G++ 4.2.1

struct Foo
{
  static const int Bar=42;
};


void SecondFunction(const int& x)
{
}

template<typename T>
void MyFunction()
{
  int x = Foo::Bar;
}

int main()
{
  MyFunction<Foo>();
  return 0;
}

compiles OK.

struct Foo
{
  static const int Bar=42;
};


void SecondFunction(const int& x)
{
}

template<typename T>
void MyFunction()
{
  SecondFunction(T::Bar);
}

int main()
{
  MyFunction<Foo>();
  return 0;
}

Gives me these errors

Undefined symbols:
“Foo::Bar”, referenced from:
void MyFunction() in cck498aS.o
ld: symbol(s) not found
collect2: ld returned 1 exit status

Answer 1

In C++, named entities fall into one of three distinct ontological tiers: values, types and templates.

If a dependent name is of an unknown tier, then you must tell the compiler what it is:

• If you say nothing, it’s a value.

• If you say typename, it’s a type.

• If you say template, it’s a template.

So T::Bar, being a dependent name inside the MyFunction template, is automatically assumed to refer to a value, hence the int member Foo::Bar. On the other hand, in the first example, typename T::Bar indeed refers to the member type Foo::Bar.