Thread: help with passing 2d array...

hi,

i just have a quick question. i have a 2d array declared like:

Code:

char array[10][6];

This array contains 10 strings. each are 5 characters long. i also have a funciton that is defined as taking an argument of type char**. Is there any way i can pass the array to this function. i've tried, but i get typing errors, and when i cast the type, the program seg faults...explinations would really help me understand this...

thank you.

Can you post you're code so we can take a look at it?

The easiest thing to do is to redefine your function to accept a 2D array, not a pointer to a pointer.

Examples

Code:

#include <stdio.h> 

void foo ( char **p ) {
  int i;
  for ( i = 0 ; i < 10 ; i++ ) {
    printf( "%s\n", p[i] );
  }
}

void bar ( char p[][5] ) {
  int i;
  for ( i = 0 ; i < 10 ; i++ ) {
    printf( "%s\n", p[i] );
  }
}

int main ( ) {
  char test[][5] = {
    "1", "2", "3", "4", "5", "6", "7", "8", "9", "10"
  };
  char *p[10];
  int i;
  for ( i = 0 ; i < 10 ; i++ ) p[i] = test[i];  // convert from [][] to **
  foo( p );
  bar ( test );
  // foo( test );	// this is a no-no
  return 0;
}

shiro, here's some sample code. The real code is too cumbersome (cryptic typdefs and such) to post, but you should get the idea from this.

Code:

#include <stdio.h>

void foo (char**);

int main ()
{
    char array[10][6] = { "fj324", "fj349", "fj239", "fd349", "fj239",
                          "fj233", "fd234", "fj239", "fj239", "fj342" };
    foo (array);  // doesn't work
    return 0;
}

void foo (char** array)
{
    int i = 0;
    while (i < 10) {
        printf ("%s\n", array[i]);
        i++;
    };
}

Salem, i just saw your reply, and you've answered my question. Thank you...however i confused as to why you can't pass the array as a pointer. Isn't a 2d array, which contains strings, the same as a pointer to a pointer. Maybe not. i just need clarification on why it doesn't work...thank you...

> Isn't a 2d array, which contains strings, the same as a pointer to a pointer
No

A 2D array is a contiguous block of memory starting at a[0][0] and finishing at a[x][y]

With a pointer to a pointer, only each level of indirection is contiguous - p[0] is followed by p[1], and p[0][0] is followed by p[0][1].
But there is no relationship between say p[0][0] and p[1][0].

Try drawing them out on paper - you'll see that they are quite different.

Although the compiler will allow you to reference either using array notation, it generates different code depending on whether the declaration is [][] or **, and this has to be consistent with what is really being passed to the function.

thank you, sir...that makes perfect sense...

you dont really have to make the function accept it, unless you need to work with every element/char in the array. just pass it like string[0] or whatever. this would be the first string of the array.