Why does the declaration void f (const int *p) is modifying p

Question

By the declaration

const int i;

it is clear that i cannot be modified.
Then why does the declaration

void f (const int *p)

is modifying p? ( I tested it, it is modifying p but don't know how?).

possible duplicate of [What is the difference between char \* const and const char \*?](http://stackoverflow.com/questions/890535/what-is-the-difference-between-char-const-and-const-char) — Oliver Charlesworth, Jul 15 '13 at 00:29
Again a wiki link: [Const Correctness](http://en.wikipedia.org/wiki/Const-correctness#Pointers_and_references) — Grijesh Chauhan, Jul 15 '13 at 05:17
@GrijeshChauhan; The sentence **Thus `const` modifies the name to its right** is confusing on the wiki link(Second para, second line). — haccks, Jul 15 '13 at 08:06
@haccks e.g in `const int *p` right of `const` is `*p`, In expression `int const * const constPtrToConst` right of first `const` after `int` is `*constPtrToConst` and right of second `const` is `constPtrToConst`. — Grijesh Chauhan, Jul 15 '13 at 08:12
@GrijeshChauhan; How `const` modify name to its right? It should not. As in example your example right of second `const` is `constPtrToConst` and it will not modify. — haccks, Jul 15 '13 at 08:40

score 7 · Accepted Answer · answered Jul 15 '13 at 00:25

The placement of const in a pointer declaration matters. You read it backwards:

const int *p means "p is a pointer to int which is constant"
int * const p means "p is a constant pointer to int"
const int * const p means "p is a constant pointer to a constant int"

Vlad · Answer 2 · 2013-07-15T23:35:49.000

Because const is int pointed to by p, not p -- the pointer to the int.

const int *p

means that p is a pointer to const int, p can be modified, *p not.

The same way

int *const p

means that p cannot be modified, whereas *p can.

Note that const int* p is the same as int const* p.

The simple rule is that the declaration reads from right to the left: int const *p means "p is a pointer to constant int", int *const p means "p is a constant pointer to an int".

(The actual complete rule is more complicated, you can use http://cdecl.org/ for "decoding" C-style declarations.)

Aniket Inge · Answer 3 · 2013-07-15T00:39:24.427

because in this expression, p is a pointer to a const int. Which means you can change what p points to.

This also means that, "p" can itself be modified, but the contents of "p" cannot be modified, so *p = 10; will yield error.

An example will clear things:

#include <stdio.h>

void foo(const int *p)
{
  int x = 10;
  p = &x; //legally OK
  //  *p = 20; <- Results in an error, cannot assign to a read-only location. 
  printf("%d\n",*p);
}
int main(int argc, char *argv[])
{
  const int x10 = 100;
  foo(&x10);
  return 0;
}

To make the above program not modify the pointer at all:

#include <stdio.h>

void foo(const int *const p)
{
  int x = 10;
  p = &x; //legally NOT OK, ERROR!
  //  *p = 20; <- Results in an error, cannot assign to a read-only location. 
  printf("%d\n",*p);
}
int main(int argc, char *argv[])
{
  const int x10 = 100;
  foo(&x10);
  return 0;
}

Why does the declaration void f (const int *p) is modifying p

3 Answers3