array of constant pointers

Question

Ok, I know that this is invalid

 char char_A = 'A';
    const char * myPtr = &char_A;
    *myPtr = 'J'; // error - can't change value of *myP

[Because we declared a pointer to a constant character]

Why is this valid?

 const char  *linuxDistro[6]={ "Debian", "Ubuntu", "OpenSuse", "Fedora", "Linux Mint", "Mandriva"};

for ( int i=0; i < 6; i++) 
cout << *(linuxDistro+i)<< endl;

*linuxDistro="WhyCanIchangeThis";// should result in an error but doesnt ? 
for ( int i=0; i < 6; i++) 
cout << *(linuxDistro+i)<< endl;

Thanks for looking!

So now is this C or C++?

user529758
– user529758

2012-11-13 06:38:05 +00:00
Commented Nov 13, 2012 at 6:38 — user529758
– user529758, Commented Nov 13, 2012 at 6:38
c++ my bad. I removed the C tag.

Illusionist
– Illusionist

2012-11-13 06:39:14 +00:00
Commented Nov 13, 2012 at 6:39 — Illusionist
– Illusionist, Commented Nov 13, 2012 at 6:39
thanks everyone for your replies, you guys are amazing!

Illusionist
– Illusionist

2012-11-13 06:51:02 +00:00
Commented Nov 13, 2012 at 6:51 — Illusionist
– Illusionist, Commented Nov 13, 2012 at 6:51

user529758 · Accepted Answer · 2012-11-13 06:41:16Z

12

You write

*linuxDistro =  "WhyCanIchangeThis";

which is perfectly valid, because the declaration of linuxDistro was

const char *linuxDistro[6];

i. e. it's an array of 6 pointers to const char. That is, you can change the pointers themselves, just not the characters pointed to by those pointers. I. e., you can not compile

*linuxDistro[0] = 'B';

to obtain the string "Bebian", becuse the strings contain constant characters...

What you probably want is an array of constant pointers to constant characters:

const char *const linuxDistro[6];

answered Nov 13, 2012 at 6:41

user529758

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7 Comments

Ashwin Over a year ago

isn't *linusDistro and *linuxDistro[0] the same?

user529758 Over a year ago

@Ashwin No. Why would it be?

Ashwin Over a year ago

the name of the array holds the memory location of the first element of the array.

Ashwin Over a year ago

I am not trying to teach you C, first of all. I was just trying to learn. You sound like a man with a big ego.

user529758 Over a year ago

@Ashwin I deleted my comment, but please... don't do this here. There are plenty of papers on this topic. This a far bigger theme than being suited for a comment.

|

Matt · Accepted Answer · 2012-11-13 06:44:19Z

3

*linuxDistro is still a pointer, it is linuxDistro[0], *linuxDistro="WhyCanIchangeThis" it just change the pointer to point to a new address, not to modify the content in the old address, so it is OK.

If you write **linuxDistro='a', it should error.

answered Nov 13, 2012 at 6:44

Matt

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Comments

user1797612 · Accepted Answer · 2012-11-13 06:45:10Z

1

because the pointer is a variable that stores a memory address, if a pointer is const the pointer keeps storing the same memory address, so the value of the pointer itself can't change, but you are saying nothing about the value pointed by the pointer, according to what you have, chaging the pointed value it's an allowed operation.

answered Nov 13, 2012 at 6:45

user1797612

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cnd · Accepted Answer · 2012-11-13 06:46:31Z

1

because char is not char[], so when you access to * of char[] you access the first element of it (Debian).

when you shift pointer (e.g. +1 it) you access next element of array

here is good example for better understanding

#include <iostream>
using namespace std;

int main ()
{
  int numbers[5];
  int * p;
  p = numbers;  *p = 10;
  p++;  *p = 20;
  p = &numbers[2];  *p = 30;
  p = numbers + 3;  *p = 40;
  p = numbers;  *(p+4) = 50;
  for (int n=0; n<5; n++)
    cout << numbers[n] << ", ";
  return 0;
}

this will output:

10, 20, 30, 40, 50,

edited Nov 13, 2012 at 6:46

answered Nov 13, 2012 at 6:40

cnd

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