C++ reference to volatile object - reasons and effects

Question

I'm working on existing C++ code that has volatile references to objects

volatile vClass & vobj;

I come from C so I'm familiar with volatile when used to access memory-mapped IO like this:

*((volatile unsigned int *) 0x12345678) = 0x5

QUESTION

What is the effect of applying volatile to (a reference to) an object?

I would guess that all of its data members inherit volatile but what about if a member function has non-volatile memory accesses like

void vClass::AccessMem() {*((unsigned int *) 0x12345678) = 0x5;}

Would that memory access too become volatile?

Answer 1

The member function must be volatile qualified to be called from a volatile object:

int not_a_member;

struct vClass {
  int i;
  void set(int j) {
     i=j; //i is not accessed as a volatile
     static_assert(std::is_same_v<decltype((i)),int&>);
     }
  void set_v(int j) volatile{
     i=j; //here i is accessed as a volatile 
     static_assert(std::is_same_v<decltype((i)),volatile int&>);
     not_a_member=j;//here not a volatile access because "not_a_member" is
                    // not a member.
     //To understand what happen, since you are a C programmer it is going to be simple.
     //The volatile qualifier is actualy apply to the "this" pointer, 
     // which is a pointer to the object from which this member function is
     // called. So inside this function "this" as the type "volatile vClass"
     //Inside a member function, all access to member data, as "i" are
     //actualy short ends for "this->i". So in this access, "i" 
     //adopt the volatile qualifier of "this".
     //So the volatile qualifier just applies to the data member of the object.
     }
  }

void use_vClass(){
   volatile vClass x;
   x.set(10); //Do not compile: "Error:try to access volatile object as non volatile"
   x.set_v(10); //Compile
  }

So since from a volatile object or reference, you can just call volatile qualified member function, all data member access will be "volatile".