Can a macro be used for read-only access to a variable?

Question

Can you define a macro that accesses a normal variable, but in a read-only fashion (other than defining it as a call to a function)? For example, can the VALUE macro in the following code be defined in such a way that the dostuff() function causes a compile error?

struct myobj {
  int value;
}

/* This macro does not satisfy the read-only requirement */
#define VALUE(o) (o)->value

/* This macro uses a function, unfortunately */
int getvalue(struct myobj *o) { return o->value; }
#define VALUE(o) getvalue(o)

void dostuff(struct myobj *foo) {
   printf("The value of foo is %d.\n", VALUE(foo)); /* OK */
   VALUE(foo) = 1; /* We want a compile error here */
   foo->value = 1; /* This is ok. */
}

Answer 1

If the variable is always numeric, this works:

#define VALUE(x) (x+0)

or in the context of your example,

#define VALUE(x) (x->value+0)

Answer 2

Ok, I came up with one:

#define VALUE(o) (1 ? (o)->value : 0)

Answer 3

See §6.5.17 in the C standard (C99 & C1x): “A comma operator does not yield an lvalue.”

#define VALUE(x) (0, x)

(Not portable to C++.)

Answer 4

Try

#define VALUE(o) (const int)((o)->value)

Answer 5

Is this a puzzle or is it an engineering task? If it's an engineering task, then there are better ways to get opacity of structures in C. In this blog article, I wrote a decent enough description of how to do that in C.