c++ redefine variable as constant

Question

I have a struct:

struct s
{
    UINT_PTR B_ID;
};
s d;
d.B_ID=0x1;

That works fine, but I want d.B_ID to be constant. I tried to use (const) but it didn't work. So after I put a value to d.B_ID, then I want make it a constant.

Any ideas?

---

EDIT

ok i don't want the whole struct a constant.
when i set timer and use the b.B_ID as an idea for the timer.
in the switch(wparam) { case b.B_ID // error: B_ID must be constant .... break; } so that is why i need it to be a constant

Answer 1

Variable modifiers are fixed at compile time for each variable. You may have to explain the context of what you are trying to do, but perhaps this will suit your needs?

struct s
{
  int* const B_ID;
};

int main (void) {
  int n = 5;
  s d = {&n};
  int* value = d.B_ID; // ok
  //  d.B_ID = &n; // error
  return 0;
}

Since you are using C++ I would recommend:

class s {
public:
    int* const B_ID;

    s (int* id) :
    B_ID (id) {
    }
};

void main (void) {
   int n = 5;
   s my_s_variable = s(&n);

   int* value = my_s_variable.B_ID; // ok
   //my_s_variable.B_ID = &n; // error
   return 0;
}

Ramiz Toma: well i need way to do it using the s.B_ID=something

In C/C++ type modifiers (like const) are declared at run time for a given type and cannot be changed at run time. This means that if a variable is declared const it can never be assigned to using the assignment operator. It will only be assigned a value when it is constructed.

This is not a problem however because you can always get around this by proper design.

If you say you need to use assignment, I assume that this is because you create the struct before you know what the value of the variable will be. If this is the case then you simply need to move the struct declaration till after you know the value.

For example

s d; //variable declaration

//calculate B_ID
//...
int* n = 5;
//...


d.B_ID = &n;

This will not work, because if you want b.D_ID to be 'un assignable' it will always be so. You will need to refactor your code similarly to:

//calculate B_ID
//...
int* n = 5;
//...


s d (&n);
//good

Answer 2

struct s
{
   s() : B_ID(0){}
   UINT_PTR const B_ID;
};
int main(){
   s d;
   d.B_ID=0x1;  // error
}

EDIT: Sorry, here is the updated code snippet in C++

struct s
{
   s(UINT_PTR const &val) : B_ID(val){}
   UINT_PTR const B_ID;
};
int main(){
   s d(1);
   d.B_ID=0x1;  // error
}

Answer 3

You can't do that - ie. it is not possible to selectively make a single member of a struct const. One option is to 'constify' the entire struct:

s d;
d.B_ID=0x1;
const s cs = s; // when using this B_ID won't be modifiable - but nor would any other members

Or you could set it at construction:

struct s
{
    s(UINT_PTR const p): B_ID(p) {}
    UINT_PTR const B_ID;
};    
s d(0xabcdef);

Answer 4

Another way would be a getter and a one time setter

class s
{
private:
   bool m_initialized;
   UINT_PTR m_value;
public:
   s() : m_initialized(false), m_value(NULL) {}
   s(UINT_PTR value) : m_initialized(true), m_value(value) {}

   //no need for copy / assignment operators - the default works

   inline UINT_PTR GetValue() const { return m_value; } //getter

   bool SetValue(UINT_PTR value) //works only one time
   {
       if (m_initialized)
       {
          m_value = value;
          m_initialized=true;
          return true;
       }
       else
       {
          return false;
       }
   }

   inline bool IsInitialized() const { return m_initialized; }
};

Answer 5

In C++ language the case label must be built from an Integral Constant Expression (ICE). ICE is what the compiler implies under the term "constant" in your error message. A non-static member of a class cannot be used in an ICE. It is not possible to do literally what you are trying to do. I.e. it is not possible to use a struct member in a case label.

Forget about switch/case in this context. Use ordinary if branching instead of switch statement.