EN VI

Why is it possible to change the value of an integer indirectly via a pointer to a constant int?

I read many old questions, many answers about const keyword in a declaration to deeply understand the syntax of declarations.

I played with const and I am now a little bit confused with this example :

int i;
i=1;
int const *ptr_i;
ptr_i=&i;
printf("%d\n",*ptr_i);
i=2;
printf("%d\n",*ptr_i);

If int const * is a pointer to a constant integer, why ptr_i=&i; is allowed ? i is not a constant integer.

Moreover, if the value of i is changed from 1 to 2 (obviously, it is possible because i is not a constant integer), the second printf displays 2. I am not surprised because the value of i has been changed and ptr_i is a pointer to i.

I, also checked that he value of ptr_i has not changed ( printf("%p ",ptr_i) ).

Where is my misunderstanding ?

Solution:

If int const * is a pointer to a constant integer,…

int const * is not a pointer to a constant integer. It is a pointer to an int for which the int type is qualified with const.

const is a misnomer. “Read only” is closer to an accurate description; it means that “with this type, the value will only be read.” Meaning that somebody else with a different pointer (same value, different type) can write to the object.

It is not purely read-only either. In C, if an object was defined without const, but you have a pointer to it of type const int *, you can convert the pointer to int * and use the new pointer to modify the object, and that is defined by the C standard. const was added to the language late, so it is somewhat grafted on rather than fully integrated, and it does not serve as a complete solution for preventing writes to objects. Mostly, it performs an advisory service: When a type is qualified with const, the compiler must warn you about attempts to use it directly to modify an object it points to. That helps programmers avoid bugs.