'ClassName' does not name a type

Question

I have seen similar answers, but I don't seem to be able to solve mine just by looking at these for example ( this or that one for example).

So, I am having that .

A.h

#ifndef INCLUDE_CLASS_NAME
#define INCLUDE_CLASS_NAME

#include <B.h>

using namespace C;

D::DPtr myvariable;  <-- Error in here

#endif

In the include B.h I have this :

namespace C{
namespace E{

class D
{
  public:
      typedef shared_ptr<D> DPtr;
}

} //end of namespace E
} // end of namespace C

Why am I getting this error in the mentioned line :

'D'  does not name a type

I am including the .h file, which defines the class. What am I missing?

Why you declared `myvariable` as `mutable`? This concept only applies to class members. — Mikhail, Jun 12 '15 at 08:55
@Mikhail yes changed that, I was not able to compile up to that point before. — ghostrider, Jun 12 '15 at 09:00

Emil Laine · Accepted Answer · 2015-06-12T08:50:56.183

5

The symbol D is inside namespace E, which is inside namespace C, so the fully qualified name is C::E::D.

So either:

edited Jun 12 '15 at 08:50

answered Jun 12 '15 at 08:44

Emil Laine

score 4 · Answer 2 · answered Jun 12 '15 at 08:43

4

You missed namespace E...

mutable E::D::DPtr myvariable; // should work

answered Jun 12 '15 at 08:43

marom

score 0 · Answer 3 · answered Jun 22 '15 at 02:33

You're trying to call a function that does not exist yet according to the preprocessor. You need to define functions before you call them:

int mainFunction(){
    int foo(int bar){ //define function FIRST...
        return bar;
    }
foo(42) //call it later.
}

This should get rid of the error. This does not:

int mainFunction(){
    foo(42) //trying to call a function before it's declared does not work.
        int foo(int bar){
        return bar;
    }
}

3 Answers3