C: Size of two dimensional array

Question

I need some help counting the rows and columns of a two dimensional array. It seems like I can\'t count columns?

#include 

int main() {

char res         


        

Answer 1

This works for me (comments explains why):

#include <stdio.h>

int main() {

   char result[10][7] = {

       {'1','X','2','X','2','1','1'},
       {'X','1','1','2','2','1','1'},
       {'X','1','1','2','2','1','1'},
       {'1','X','2','X','2','2','2'},
       {'1','X','1','X','1','X','2'},
       {'1','X','2','X','2','1','1'},
       {'1','X','2','2','1','X','1'},
       {'1','X','2','X','2','1','X'},
       {'1','1','1','X','2','2','1'},
       {'1','X','2','X','2','1','1'}

   }; 

   // 'total' will be 70 = 10 * 7
   int total = sizeof(result);

   // 'column' will be 7 = size of first row
   int column = sizeof(result[0]);

   // 'row' will be 10 = 70 / 7
   int row = total / column;

   printf("Total fields: %d\n", total);
   printf("Number of rows: %d\n", row);
   printf("Number of columns: %d\n", column);

}

And the output of this is:

Total of fields: 70
Number of rows: 10
Number of columns: 7

EDIT:

As pointed by @AnorZaken, passing the array to a function as a parameter and printing the result of sizeof on it, will output another total. This is because when you pass an array as argument (not a pointer to it), C will pass it as copy and will apply some C magic in between, so you are not passing exactly the same as you think you are. To be sure about what you are doing and to avoid some extra CPU work and memory consumption, it's better to pass arrays and objects by reference (using pointers). So you can use something like this, with same results as original:

#include <stdio.h>

void foo(char (*result)[10][7])
{
   // 'total' will be 70 = 10 * 7
   int total = sizeof(*result);

   // 'column' will be 7 = size of first row
   int column = sizeof((*result)[0]);

   // 'row' will be 10 = 70 / 7
   int row = total / column;

   printf("Total fields: %d\n", total);
   printf("Number of rows: %d\n", row);
   printf("Number of columns: %d\n", column);

}

int main(void) {

   char result[10][7] = {

       {'1','X','2','X','2','1','1'},
       {'X','1','1','2','2','1','1'},
       {'X','1','1','2','2','1','1'},
       {'1','X','2','X','2','2','2'},
       {'1','X','1','X','1','X','2'},
       {'1','X','2','X','2','1','1'},
       {'1','X','2','2','1','X','1'},
       {'1','X','2','X','2','1','X'},
       {'1','1','1','X','2','2','1'},
       {'1','X','2','X','2','1','1'}

   };

   foo(&result);

   return 0;
}

Answer 2

Use the macros shown in below code to get any dimension size of 1D, 2D or 3D arrays. More macros can be written similarly to get dimensions for 4D arrays and beyond. (i know its too late for Wickerman to have a look but these're for anyone else visiting this page)

// Output of the following program
// [
/*

Demo of the advertised macros :
----------------------------------------------
sizeof(int) = 4
sizeof(Array_1D) = 12
ELEMENTS_IN_1D_ARRAY(Array_1D) = 3
sizeof(Array_2D) = 24
ELEMENTS_IN_2D_ARRAY(Array_2D) = 6
ROWS_IN_2D_ARRAY(Array_2D) = 2
COLUMNS_IN_2D_ARRAY(Array_2D) = 3
sizeof(Array_3D) = 96
ELEMENTS_IN_3D_ARRAY(Array_3D) = 24
MATRICES_IN_3D_ARRAY(Array_3D) = 4
ROWS_IN_3D_ARRAY(Array_3D) = 2
COLUMNS_IN_3D_ARRAY(Array_3D) = 3

Array_3D[][][] Printed :
----------------------------------------------
 001 002 003
 011 012 013
---------------
 101 102 103
 111 112 113
---------------
 201 202 203
 211 212 213
---------------
 301 302 303
 311 312 313
---------------

Wickerman's problem solved :
----------------------------------------------
sizeof(result) = 70
ELEMENTS_IN_2D_ARRAY(result) = 70
ROWS_IN_2D_ARRAY(result) = 10
COLUMNS_IN_2D_ARRAY(result) = 7

*/
// ]

// ====================================================================================================
// Program follows
// ====================================================================================================

// Array Size Macros
// [
#define ELEMENTS_IN_1D_ARRAY(a1D)   ( sizeof( a1D       ) / sizeof( a1D[0]          )) // Total no. of elements in 1D array
#define ELEMENTS_IN_2D_ARRAY(a2D)   ( sizeof( a2D       ) / sizeof( a2D[0][0]       )) // Total no. of elements in 2D array
#define ROWS_IN_2D_ARRAY(a2D)       ( sizeof( a2D       ) / sizeof( a2D[0]          )) // No. of Rows in a 2D array
#define COLUMNS_IN_2D_ARRAY(a2D)    ( sizeof( a2D[0]    ) / sizeof( a2D[0][0]       )) // No. of Columns in a 2D array
#define ELEMENTS_IN_3D_ARRAY(a3D)   ( sizeof( a3D       ) / sizeof( a3D[0][0][0]    )) // Total no. of elements in 3D array
#define MATRICES_IN_3D_ARRAY(a3D)   ( sizeof( a3D       ) / sizeof( a3D[0]          )) // No. of "Matrices" (aka "Slices"/"Pages") in a 3D array
#define ROWS_IN_3D_ARRAY(a3D)       ( sizeof( a3D[0]    ) / sizeof( a3D[0][0]       )) // No. of Rows in each "Matrix" of a 3D array
#define COLUMNS_IN_3D_ARRAY(a3D)    ( sizeof( a3D[0][0] ) / sizeof( a3D[0][0][0]    )) // No. of Columns in each "Matrix" of a 3D array
// ]

#define PRINTF_d(s) (printf(#s " = %d\n", (int)(s)))    // Macro to print a decimal no. along with its corresponding decimal expression string,
                                                        // while avoiding to write the decimal expression twice.

// Demo of the Array Size Macros defined above
// [
main()
{
    // Sample array definitions
    // [
    int Array_1D[3] = {1, 2, 3};    // 1D array

    int Array_2D[2][3] =            // 2D array
    {
        {1,  2,  3},
        {11, 12, 13}
    };

    int Array_3D[4][2][3] =         // 3D Array
    {
        {
            {1,   2,   3},
            {11,  12,  13}
        },
        {
            {101, 102, 103},
            {111, 112, 113}
        },
        {
            {201, 202, 203},
            {211, 212, 213}
        },
        {
            {301, 302, 303},
            {311, 312, 313}
        }
    };
    // ]

    // Printing sizes and dimensions of arrays with the advertised Array Size Macros
    printf(
    "Demo of the advertised macros :\n"
    "----------------------------------------------\n");
    PRINTF_d(sizeof(int));
    PRINTF_d(sizeof(Array_1D));
    PRINTF_d(ELEMENTS_IN_1D_ARRAY(Array_1D));
    PRINTF_d(sizeof(Array_2D));
    PRINTF_d(ELEMENTS_IN_2D_ARRAY(Array_2D));
    PRINTF_d(ROWS_IN_2D_ARRAY(Array_2D));
    PRINTF_d(COLUMNS_IN_2D_ARRAY(Array_2D));
    PRINTF_d(sizeof(Array_3D));
    PRINTF_d(ELEMENTS_IN_3D_ARRAY(Array_3D));
    PRINTF_d(MATRICES_IN_3D_ARRAY(Array_3D));
    PRINTF_d(ROWS_IN_3D_ARRAY(Array_3D));
    PRINTF_d(COLUMNS_IN_3D_ARRAY(Array_3D));

    // Printing all elements in Array_3D using advertised macros
    // [
    int x, y, z;

    printf(
    "\nArray_3D[][][] Printed :\n"
    "----------------------------------------------\n");

    for(x = 0; x < MATRICES_IN_3D_ARRAY(Array_3D); x++)
    {
        for(y = 0; y < ROWS_IN_3D_ARRAY(Array_3D); y++)
        {
            for(z = 0; z < COLUMNS_IN_3D_ARRAY(Array_3D); z++)
                printf("%4.3i", Array_3D[x][y][z]);
            putchar('\n');
        }
        printf("---------------\n");
    }
    // ]

    // Applying those macros to solve the originally stated problem by Wickerman
    // [
    char result[10][7] = {
        {'1','X','2','X','2','1','1'},
        {'X','1','1','2','2','1','1'},
        {'X','1','1','2','2','1','1'},
        {'1','X','2','X','2','2','2'},
        {'1','X','1','X','1','X','2'},
        {'1','X','2','X','2','1','1'},
        {'1','X','2','2','1','X','1'},
        {'1','X','2','X','2','1','X'},
        {'1','1','1','X','2','2','1'},
        {'1','X','2','X','2','1','1'}
    };

    printf(
    "\nWickerman's problem solved :\n"
    "----------------------------------------------\n");
    PRINTF_d(sizeof(result)); // radha_SIZEOF_2D_ARRAY
    PRINTF_d(ELEMENTS_IN_2D_ARRAY(result)); // radha_SIZEOF_2D_ARRAY
    PRINTF_d(ROWS_IN_2D_ARRAY(result));
    PRINTF_d(COLUMNS_IN_2D_ARRAY(result));
    // ]
}
// ]

Answer 3

It's much more convenient (and less error prone) to use an array length macro:

#include <stdio.h>

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

int main(void)
{
    char result[10][7];

    printf("Number of rows: %d\n", LEN(result));
    printf("Number of columns: %d\n", LEN(result[0]));
    return 0;
}

Answer 4

    // gets you the total size of the 2d array 
    printf("Arrays Total size: %ld\n",sizeof(result));

    // gets you the cumulative size of row which is 5 columns * sizeof(int)
    printf("1 row cumulative size: %ld\n",sizeof(result[0]));

    // division of total array size with cumulative size of row gets you total number of rows
    printf("total number of rows: %ld\n",sizeof(result)/sizeof(result[0]));

    // and total number of columns you get by dividing cumulative row size with sizeof(char)
    printf("total number of columns: %ld\n",sizeof(result[0])/sizeof(char));

Answer 5

That's a problem of integer division!

int column = sizeof(result[0])/row;

should be

int column = 7 / 10;

and in integer division, 7/10==0.

What you want to do is divide the length of one row, eg. sizeof(result[0]) by the size of one element of that row, eg. sizeof(result[0][0]):

int column = sizeof(result[0])/sizeof(result[0][0]);