Dereferencing a pointer to constant

Question

Let's say we have a class called object.

int main(){     object a;     const object* b = &a;     (*b);  } 

Question: b is a pointer to const, but the object it is pointing to is actually not a constant object. My question is, when we dereference the pointer "b" why do we get a constant object instead of what it is actually pointing at which is normal non constant object.

Answer 1

Because that's how the built-in dereference operator * works in C++. If you dereference a pointer of type T *, you get an lvalue of type T. In your case T is const object.

Neither * operator, not the pointer itself cares (or knows) that the object the pointer is pointing to is actually non-constant. It just can't be any other way within the concept of static typing used by C++ language.

The whole purpose of const-qualification at the first (and deeper) levels of indirection is to provide you with an ability to create restrictive access paths to objects. I.e. by creating a pointer-to-const you are deliberately and willingly preventing yourself (or someone else) from modifying the pointee, even if the pointee is not constant.

Answer 2

The type of an expression is just based on the declared types of the variables in the expression, it can't depend on dynamic run-time data. For instance, you could write:

object a1; const object a2; const object *b; if (rand() % 2 == 0) {     b = &a1; } else {     b = &a2; } (*b); 

The type of *b is determined at compile-time, it can't depend on what rand() returns when you run the program.

Since b is declared to point to const object, that's what the type of *b is.