Initializing 2D char array in C

Question

Similar to this question: 2d array, using calloc in C

I need help initializing a 2D char array that will all be initialized to some value (in this case '0'). I have tried many different methods and I am pulling my hair out. Please let me know what I am doing wrong. This code doesn't work. Thanks!

char** init_array() {
        char newarray[5][10];
        int i, j;
        for (i = 0; i < 5; i++) {
                for (j = 0; j < 10; j++) {
                        newarray[i][j] = '0';
                }
        }
        return newarray;
}
char **array = init_array();

The errors I get from gcc when I try to compile:

test.c: In function ‘init_array’:
test.c:12:2: warning: return from incompatible pointer type [enabled by default]
  return newarray;
  ^
test.c:12:2: warning: function returns address of local variable [-Wreturn-local-addr]
test.c: At top level:
test.c:14:1: error: initializer element is not constant
 char **array = init_array();

Should it be like this?

char newarray[5][10];
char** init_array() {
        int i, j;
        for (i = 0; i < 5; i++) {
                for (j = 0; j < 10; j++) {
                        newarray[i][j] = '0';
                }
        }
        return newarray;
}
char **array = init_array();

You can't return a local array, you need to allocate it dynamically with `malloc`. — Barmar, Nov 15 '14 at 01:25
Or pass the pointer as an argument and then you can (1) operate on it in the function (void type behavior) and (2) return a value (char** behavior) — David C. Rankin, Nov 15 '14 at 01:26
newarray goes out of scope once you return from the function. Allocate memory inside init_array. — helloV, Nov 15 '14 at 01:27
Really? I wasn't aware of that. I thought they were the same. — mallux, Nov 15 '14 at 01:28
No-- that is your error: `return from incompatible pointer type`. `char**` is a pointer to a character pointer, `char[][]` is an array of character arrays. — David C. Rankin, Nov 15 '14 at 01:29
possible duplicate of [Return a 2d array from a function](http://stackoverflow.com/questions/8617889/return-a-2d-array-from-a-function) — M.M, Nov 15 '14 at 04:09
Your initializing is OK, the problem is with how you try to return the array — M.M, Nov 15 '14 at 04:10

Mike Dunlavey · Accepted Answer · 2014-11-15T02:21:10.540

19

I think pictures help. Here is char newarray[5][10]. It is a single memory block consisting of an array of 10 characters, and an array of five of those. You could just clear it with a single memset call.

enter image description here

Here is char **array. It says array is a pointer. What is it a pointer to? a pointer to a character.

enter image description here

Keep in mind pointer arithmetic. If array is a pointer that happens to point to a pointer, then (*array) equals array[0], and that is the pointer that array points to.

What is array[1]? It is the second pointer in the array that array points to.

What is array[0][0]? It is the first character pointed at by the first pointer that array points to.

What is array[i][j]? It is the jth character of the ith pointer that array points to.

So how are newarray and array related?
Simple. newarray[i][j] is the jth character of the ith subarray of newarray.

So in that sense, it's just like array, but without all the pointers underneath.

What's the difference? Well, the disadvantage of array is you have to build it up, piece by piece. OTOH, the advantage is you can make it as big as you want when you build it up. It doesn't have to live within a fixed size known in advance.

Clear as mud?

edited Nov 15 '14 at 02:21

answered Nov 15 '14 at 02:07

Mike Dunlavey

40,059
14
91
135

Thanks for explaining that to me and letting me know about memset(). I'm new to C... – mallux Nov 15 '14 at 02:21
3

You can also simply zero a statically declared array when you declare it: `char newarray[5][10] = {{0}};` – David C. Rankin Nov 15 '14 at 02:24
That does not work. I just tried it. I wish it was that easy! – mallux Nov 15 '14 at 02:27
3

Hold on, you need to `#define` the sizes not provide them as if creating a variable length array. I'll post a supplemental answer for you. – David C. Rankin Nov 15 '14 at 02:30
2

@mallux `char newarray[5][10] = { 0 };` – M.M Nov 15 '14 at 02:33
@MattMcNabb gcc will warn and suggest additional braces :) `warning: missing braces around initializer` – David C. Rankin Nov 15 '14 at 02:36

score 3 · Answer 2 · answered Nov 15 '14 at 02:34

Per our discussion in the comments, here is a quick example of zeroing array values at the time of declaration. Note, the values are #defined as constants:

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

#define MAXSTR 10
#define MAXLEN 1024

int main () {

    char myarray[MAXSTR][MAXLEN] = {{0}};  /* declare a static array of MAXSTR x MAXLEN */

    /* copy various strings to statically declared storage */
    strncpy (myarray[0],"  This is the first string", strlen("  This is the first string")+1);
    strncpy (myarray[1],"  This is the second string", strlen("  This is the second string")+1);
    strncpy (myarray[2],"  This is the third string", strlen("  This is the third string")+1);
    strncpy (myarray[3],"  This is the fourth string", strlen("  This is the fourth string")+1);
    strncpy (myarray[4],"  This is the fifth string", strlen("  This is the fifth string")+1);

    int i = 0;

    /* print the values */
    while (*myarray[i])
        printf ("  %s\n", myarray[i++]);

    return 0;
}

output:

$ ./myar
    This is the first string
    This is the second string
    This is the third string
    This is the fourth string
    This is the fifth string

build:

gcc -Wall -Wextra -o myar myarray.c

Yeah you're right. It does "zero" them but it doesn't actually set them to '0'. That's what I thought you meant. — mallux, Nov 15 '14 at 02:50
Sure, do you just want an example program such as reading a file into a dynamically allocated array or array of structures? — David C. Rankin, Jun 08 '15 at 08:25

twasbrillig · Answer 3 · 2014-11-15T01:51:33.623

To avoid using a global (like the second sample code pasted above) and to avoid using malloc, you can define the array outside your function and pass it in, like this. You don't need to return anything because the array data itself is being modified. Notice that it's necessary to define the array's secondary dimension in the function signature:

void init_array(char ary[][10]) {
    int i, j;
    for (i = 0; i < 5; i++) {
        for (j = 0; j < 10; j++) {
                ary[i][j] = '0';
        }
    }
}

int main(void)
{
    char newarray[5][10];
    init_array(newarray);
    printf("%c", newarray[1][1]); /* Testing the output */
    return 0;
}

That returns '0'.

http://codepad.org/JbykYcyF

Make sure you pass (or #define) the sizes so you are not hardcoding values in the fuction: `void init_array(char ary[][10], int rows)` — David C. Rankin, Nov 15 '14 at 02:27

Initializing 2D char array in C

3 Answers3