Store multidimensional array in struct

Question

I have a two dimensional array where the first dimension has a variable length but the second dimension is fixed. Now in a function call I could do something like char foo[][3] but what is the corresponding definition in a struct?

So in the example code I expected it to print each string in a line, but as expected it treats the stored pointer as a single dimensional array.

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

struct payload_s {
    size_t length;
    char *text;
};

typedef struct payload_s payload;

static char some_text[4][3] = {"Ab", "Cd", "Ef", "Gh"};

payload* create_payload(char *text, size_t length)
{
    payload *p = malloc(sizeof *p);
    p->text = text;
    p->length = length;
    return p;
}

int main()
{
    payload *p = create_payload(some_text, 4);
    for (size_t i = 0; i < p->length; ++i)
        printf("%zu: %s\n", i, &p->text[i]);
}

I mainly noticed this because of a warning:

strut.c: In function ‘main’:
strut.c:23:33: warning: passing argument 1 of ‘create_payload’ from incompatible pointer type [-Wincompatible-pointer-types]
     payload *p = create_payload(some_text, 4);
                                 ^~~~~~~~~
strut.c:13:10: note: expected ‘char *’ but argument is of type ‘char (*)[3]’
 payload* create_payload(char *text, size_t length)
          ^~~~~~~~~~~~~~

I can get rid of this warning when the function is actually defined as payload* create_payload(char text[][3], size_t length), but then there is a warning a few lines later and the behavior didn't change:

strut.c: In function ‘create_payload’:
strut.c:16:13: warning: assignment from incompatible pointer type [-Wincompatible-pointer-types]
     p->text = text;
             ^

Is the only solution to manually increment the pointer by the length of each value?

You have conceptual issue: You cannot store four char arrays ({"Ab\0", "Cd\0", "Ef\0", "Gh\0"}; here) under one char-pointer (char *text; here) only, at least not in sane way. The compiler warnings are a hint to this issue. — alk
– alk, Commented Nov 16, 2016 at 17:16
That is basically what I'm asking: How else can I store them in a struct? — xZise
– xZise, Commented Nov 16, 2016 at 17:21
OT: "Ab\0" is represented internally as {'A', 'b', '\0', '\0'}, so you probably do not want to place the '\0' explicitly. — alk
– alk, Commented Nov 16, 2016 at 17:21
Change the member char * text to be a char (*)[3] as well: Make it char (*text)[3]. — alk
– alk, Commented Nov 16, 2016 at 17:26
when ever calling any of the heap memory allocation functions: (malloc, calloc, realloc), always check (!=NULL) the returned value to assure the operation was successful — user3629249
– user3629249, Commented Nov 16, 2016 at 17:31

Community · Accepted Answer · 2017-05-23 12:13:44Z

1

Use:

char (*text)[3];

instead of:

char *

since what you want here is a pointer to your 2D array, not a pointer to single char. Read more in C pointer to two dimensional array.

Of course, it is suggested to use a define, or something similar for your dimension, instead of the hardcoded 3, like in my example.

Min. Example:

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

#define M 3

struct payload_s {
    size_t length;
    char (*text)[M]; // change the member!
};

typedef struct payload_s payload;

// not need for null terminators in the strings,
// it will be placed automatically
static char some_text[4][M] = {"Ab", "Cd", "Ef", "Gh"};

// change the prototype as well
payload* create_payload(char (*text)[M], size_t length)
{
    payload *p = malloc(sizeof *p);
    p->text = text;
    p->length = length;
    return p;
}

int main()
{
    payload *p = create_payload(some_text, 4);
    for (size_t i = 0; i < p->length; ++i)
        // no need to print the address now
        // also 'zu' should be used for 'size_t'
        printf("%zu: %s\n", i, p->text[i]);
}

Output:

Georgioss-MacBook-Pro:~ gsamaras$ gcc -Wall main.c 
Georgioss-MacBook-Pro:~ gsamaras$ ./a.out 
0: Ab
1: Cd
2: Ef
3: Gh

PS - Check what malloc() returns, to see if the memory was actually allocated (of course on real code, not in min. egs).

edited May 23, 2017 at 12:13

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answered Nov 16, 2016 at 17:24

gsamaras

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2 Comments

xZise Over a year ago

I almost got it but placed the brackets wrongly (used *(text[3])). Appreciate the additional comments, though this code was only to extract the issue I had.

gsamaras Over a year ago

I know @xZise, that's why my last sentence! I had to look it up myself, that's why I linked you the answer I checked before writing this answer! :) Good question by the way, very refreshing!

user3629249 · Accepted Answer · 2016-11-16 18:14:22Z

this line:

static char some_text[4][3] = {"Ab\0", "Cd\0", "Ef\0", "Gh\0"};

is trying to initialize an array of 3 bytes in each entry with values that are 4 bytes in each entry.

note: "AB\0" is 4 bytes because declaring a char array Always appends a NUL byte to the end of the array.

Suggest:

static char some_text[4][3] = {"Ab", "Cd", "Ef", "Gh"};

The call to printf() contains several errors, which your compiler should have told you about.

The field: char *text; is actually pointing to a 2D array, so should be declared accordingly.

Need to perform error checking on the call to malloc().

here is a version of the code, will all (reasonable) corrections applied.

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

struct payload_s 
{
    size_t length;
    char **text;
};


static char *some_text[] = {"Ab", "Cd", "Ef", "Gh"};

struct payload_s* create_payload(char **text, size_t length)
{
    payload *p = malloc(sizeof (struct payload_s));
    if( !p )
    {
        perror( "malloc for instance of payload failed" );
        exit( EXIT_FAILURE );
    }

    // implied else, malloc successful

    p->text = text;
    p->length = length;
    return p;
}

int main( void )
{
    //payload *p = create_payload(some_text, 4);
    payload *p = create_payload(some_text, sizeof(some_text) / sizeof( *some_text ) );
    for (size_t i = 0; i < p->length; ++i)
        printf("%lu: %s\n", i, p->text[i]);
}

The result of the above code is:

0: Ab
1: Cd
2: Ef
3: Gh

Vlad from Moscow · Accepted Answer · 2016-11-16 18:23:34Z

You are using incompatible pointers that can not be converted implicitly to each other.

Character array some_text declared like

static char some_text[4][3] = {"Ab\0", "Cd\0", "Ef\0", "Gh\0"};

when it is used in expression as for example used as an argument it is implicitly converted to pointer to its first element and has type char ( * )[3]. It si not the same as pointer of type char *.

It looks like you need a structure with a flexible array.

Here is a demonstrative program that shows how such a structure can be used.

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

#define N   3

struct payload_s 
{
    size_t length;
    char text[][N];
};

struct payload_s * create_payload( char( *text )[N], size_t length )
{
    struct payload_s *p = malloc( sizeof( struct payload_s ) + 
                                  length * sizeof( char[N] ) );

    p->length = length;

    for ( size_t i = 0; i < length; i++ ) strcpy( p->text[i], text[i] );

    return p;
}

void free_payload( struct payload_s *p )
{
    free( p );
}

int main(void) 
{
    char some_text[4][N] = {"Ab", "Cd", "Ef", "Gh"};
    char another_text[5][N] = {"Bb", "Dd", "Ff", "Hh", "Jj"};

    struct payload_s *p1 = create_payload( some_text, 
        sizeof( some_text ) / sizeof( *some_text ) );

    struct payload_s *p2 = create_payload( another_text, 
        sizeof( another_text ) / sizeof( *another_text ) );

    for ( size_t i = 0; i < p1->length; i++ )
    {
        printf( "%s ", p1->text[i] );
    }
    printf( "\n" );

    for ( size_t i = 0; i < p2->length; i++ )
    {
        printf( "%s ", p2->text[i] );
    }
    printf( "\n" );

    free_payload( p2 );
    free_payload( p1 );

    return 0;
}

Its output is

Ab Cd Ef Gh 
Bb Dd Ff Hh Jj

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