将二维数组分配给结构对象

Question

I have 2-dimensional array. I want that array to be assigned to struct. Please take a look:

This is my struct:

typedef struct {
    int x;
    int y;
    char **table; //2-dim array
} some_struct;

And I want to be able to assign this:

const char sth1_table[2][3] = {
   { ' ', 'x', ' '},
   { 'x', 'x', 'x'},
};

or this:

const char sth2_table[4][2] = {
   { 'x', ' '},
   { 'x', ' '},
   { 'x', ' '},
   { 'x', 'x'},
};

to that struct.

How to do that? I tried assigning:

new_sth.table = malloc(sizeof(sth1_table));
*new_sth.table = sth1_table;

And then accessing:

some_struct get_sth;
get_sth = *(some_struct+i);
other_array[a][b] = get_sth.table[a][b];

But with no luck.

What am I doing wrong?

Answer 1

For starters the member table is not a two dimensional array. It is a pointer.

Arrays do not have the assignment operator. You have to copy arrays element by element.

For example if you have an object of the structure

typedef struct {
    int x;
    int y;
    char **table; // pointer
} some_struct;

some_struct obj;

and array

const char sth1_table[2][3] = {
   { ' ', 'x', ' '},
   { 'x', 'x', 'x'},
};

then you can make the following way

obj.table = malloc( sizeof( char *[2] ) );

for ( size_t i = 0; i < 2; i++ )
{
    obj.table[i] = malloc( sizeof( sth1_table[i] ) );
    memcpy( obj.table[i], sth1_table[i], sizeof( sth1_table[i] ) );
} 

You should remember that you have to free all allocated memory or realloc it when you need to use another two-dimensional array as an initializer.

Answer 2

If you can do it with the array sizes hard coded and without using malloc, I would consider this example:

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

struct Some_Struct
{
    int x;
    int y;
    char sth1_table[2][3];
    char sth2_table[4][4];
};

int main ( int argc, char *argv[] )
{
    struct Some_Struct yy;
    struct Some_Struct zz[10];

    yy.x = 1;
    yy.y = 2;

    /*
    you wanted this
    const char sth1_table[2][3] = {
        { ' ', 'a', ' '},
        { 'b', 'c', 'd'},
    };
    */

    yy.sth1_table[0][0] = '\0';   /* null character */
    yy.sth1_table[0][1] = 'a';
    yy.sth1_table[0][2] = '\0';
    yy.sth1_table[1][0] = 'b';
    yy.sth1_table[1][1] = 'c';
    yy.sth1_table[1][2] = 'd';

    for ( i = 0; i < 10; i++ )
    {
        zz[i].sth1_table[0][0] = '\0';   /* null character */
        zz[i].sth1_table[0][1] = 'a';
        zz[i].sth1_table[0][2] = '\0';
        zz[i].sth1_table[1][0] = 'b';
        zz[i].sth1_table[1][1] = 'c';
        zz[i].sth1_table[1][2] = 'd';
    }
}

if you are interested in runtime allocation then also consider the calloc() function in addition to malloc(). When doing so, remember the memory allocation of your 2-d array (or 3-d, 4-d, any-dimension) is just a chunk of memory whose size is the number of dimensions multiplied by the type size.

char sth1_table[2][3];

the size of char is 1 byte. 1 byte * 2 * 3 = 6 bytes of memory for sth1_table

char sth2_table[4][4];

1 byte * 4 * 4 = 16 bytes of memory sth2_table

How you then chose to order your data within that allocated memory, whether it is row first or column first, is entirely up to you.

Answer 3

Instead of using malloc directly I suggest defining a macro for memory allocation. This makes the code easier to follow. In the example below I have taken the liberty to rename some identifiers.

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

#define LEN(arr) (sizeof (arr) / sizeof (arr)[0])

#define NEW_ARRAY(ptr, n) (ptr) = malloc((n) * sizeof (ptr)[0])

typedef struct {
    int m;
    int n;
    char **table;
} Matrix;


int main()
{
    const char items[2][3] = {{' ', 'x', ' '}, {'x', 'x', 'x'}};
    Matrix A;
    int i, j;

    A.m = LEN(items);
    A.n = LEN(items[0]);

    /*initialize matrix*/   
    NEW_ARRAY(A.table, A.m);
    for (i = 0; i < A.m; i++) {
        NEW_ARRAY(A.table[i], A.n);
        for (j = 0; j < A.n; j++) {
            A.table[i][j] = items[i][j];
        }
    }

    /*print matrix*/
    for (i = 0; i < A.m; i++) {
        for (j = 0; j < A.n; j++) {
            printf("%c ", A.table[i][j]);
        }
        putchar('\n');
    }

    return 0 ;
}