two-dimensional array in C: switching between pointer and index notation

frankie_DJ · 02-22-2007, 06:54 PM

Hi everyone,

I think this is an easy question, I am just a bit rusty. I want to have an array of strings, each string 3 characters long. So I declare it as

Code:

char (*name)[3];

If I am not mistaken, without parens this would be an array of 3 pointers to char, right?

OK so then I set aside some space for this array with:

Code:

name=malloc(N);

Then, when I want to assign values to the array elements, I want to switch to the index notation (it is more convinient for the rest of the code), so I write:

Code:

name[index][3]=some_short_string;

But gcc complains, saying

Code:

"warning: assignment makes integer from pointer without a cast".

Well, yeah, I can understand how it can be seen that way, but what's the right way to do it? Even when I stick with the pointer notation and do

Code:

*(name++)[3]=some_short_string;

it gives me the same warning. Thanks for your help.

frankie_DJ · 02-22-2007, 11:07 PM

Man this was wrong on so many levels. I can't really do assignements like that, have to use strcpy; and there is no need to mention the size [3] after initializing. So this compiles:

Code:

strcpy(*name,some_short_string);

But how do I switch to indices?

Dark_Helmet · 02-22-2007, 11:22 PM

I'm just tried this on my box:

Code:

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

typedef mystring[4]; /* Each string has three characters (plus one for NULL) */

int main( int argc, char *argv[] )
{
  mystring *names;

  names = malloc( sizeof(mystring) * 3 );

  strcpy( names[0], "joe" );
  strcpy( names[1], "bob" );
  strcpy( names[2], "jim" );

  printf( "%c%c%c\n", names[0][0], names[1][1], names[2][2] );

  free( names );

  return 0;
}

varun_shrivastava · 02-22-2007, 11:44 PM

for using indices u need to use loop which will copy character by character.
and the relation b/w pointers and indices is :

a[0] is same as *(a+0)
a[1] is same as *(a+1)... and so on.