array of char * converted to char *

Question

How to retrieve the data from domain_cx in domaine , once I'm in the function get_domaine(...) I have tried the following, but the result isnt what expected (0-2 instead of 0-2 3). Here is my code

#include <stdio.h>
#include <stdlib.h>

int get_domaine(char * domaine)
{
  char (*ptr_dom)[10];
  ptr_dom = (char (*)[10]) domaine;
  printf("%s ", ptr_dom[0]); // will print
  printf("%s ", ptr_dom[1]); // will not print even thought there is data.
  return 1;
}

int main(int argc,char * argv[])
{
   char *domaine_cx[10];

   domaine_cx[0] = "0-2";
   domaine_cx[1] = "3";

  if(get_domaine((char *)*domaine_cx)) printf("Ok");
  return EXIT_SUCCESS;
}

I edited it, it does compile now.

maximegir
– maximegir

2013-10-20 23:22:51 +00:00
Commented Oct 20, 2013 at 23:22 — maximegir
– maximegir, Commented Oct 20, 2013 at 23:22

LihO · Accepted Answer · 2013-10-20 23:32:23Z

5

You are trying to cast char (*) [10] to char * just so that you can try to cast it back afterwards. Much simpler would be changing the prototype of your function instead:

int get_domaine(char* domaine[10])
{
    printf("%s ", domaine[0]);
    printf("%s ", domaine[1]);
    return 1;
}

and in main:

if (get_domaine(domaine_cx))
    printf("Ok");

Alternatively you might not pass an array of any size and pass the size in other argument explicitly:

int get_domaine(char* domaine[], int tokens)
{
    int i;
    for (i = 0; i < tokens; ++i)
        printf("%s ", domaine[i]);
    return 1;
}

and in main:

if (get_domaine(domaine_cx, 2))
    printf("Ok");

And in case you really can not change the prototype, than this evil cast would do:

int get_domaine(char* domaine)
{
    char** myEvilPtr = (char**) domaine;       // !!! we don't know the size !!!
    printf("%s ", myEvilPtr[0]);
    printf("%s ", myEvilPtr[1]);
    return 1;
}

and since arrays decay into pointers, in main you could still do simple:

if (get_domaine(domaine_cx))
    printf("Ok");

edited Oct 20, 2013 at 23:32

answered Oct 20, 2013 at 23:22

LihO

42.3k11 gold badges103 silver badges169 bronze badges

Sign up to request clarification or add additional context in comments.

10 Comments

user2819245 Over a year ago

try function declaration int get_domaine(char* domaine[]), or get_domaine(char** domaine)

maximegir Over a year ago

I have no control over the signature, unfortunately

maximegir Over a year ago

Since i know the size of the array of char* I can cast it to a pointer to a char (the first char) , knowing the size , can't I use it from that pointer to a char ?

icktoofay Over a year ago

@user2827260: As far as I'm aware, there are no guarantees that a char * will be the same size as a char **, although on most systems you'll encounter that will be the case. As such, anything you can do will probably be undefined behavior.

icktoofay Over a year ago

@elgonzo: You can't cast pointers around willy-nilly. You can cast a pointer to something to a pointer to char, a pointer to void, back to whatever it originally was pointing to, and, as far as I know, nothing else without triggering undefined behavior. Furthermore, as far as I know, the standard gives no guarantees to sizes of pointers beyond that a void * is big enough to hold any other pointer.

|

Collectives™ on Stack Overflow

array of char * converted to char *

1 Answer 1

10 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

10 Comments

Your Answer

Sign up or log in

Post as a guest

Related