2d array initialized with malloc/calloc

Question

Alright, I'm having trouble understanding how to use malloc in calloc to initalize an array. I'm trying to do some practice by creating a 2 * 3 matrix that stores user-input values. The only part of the code that I don't want to change is using **matrix instead of matrix[5][7]. Any suggestions?

Here's my code so far (I keep getting segmentation faults):

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

main(){

    int i, j;

    int **mat = (int **)malloc(2 * 3 * sizeof(int*));

    for(i=0;i<2;i++)
      for(j=0;j<3;j++){

      printf("Input a value for Array[%d][%d]: ",i,j);

      scanf("%d",&mat[i][j]);

      }
    for(i=0;i<2;i++)
      for(j=0;j<3;j++)
        printf("%d\t",mat[i][j]);

}

EDIT: I appreciate the help everyone! My code now works without faults. Here's the what it looks like:

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

main(){

int i, j;

int **mat;

  mat = malloc(2 *sizeof(int *));

for(i=0;i<2;i++){

  mat[i] = malloc(3 *sizeof(int));

        for(j=0;j<3;j++){

        printf("Input a value for Array[%d][%d]: ",i,j);

        scanf("%d",&mat[i][j]);

        }
  }

for(i=0;i<2;i++)
  for(j=0;j<3;j++)
  printf("%d\t",mat[i][j]);

return 0;
}

If there's any reason I should make more edits, let me know.

Don't use implicit `int` return-type. Also, [Don't cast the result of malloc (and friends)](http://stackoverflow.com/q/605845). — Deduplicator, Dec 10 '14 at 04:59
look at "related", there are hundreds of threads for how to malloc a 2-d array. — M.M, Dec 10 '14 at 09:44
the edited version of the program still has one problem. the line 'int main( void )' will cause a compiler (with all warnings enabled) to complain. The line should be: 'int main()' — user3629249, Dec 10 '14 at 22:10

Gopi · Answer 1 · 2014-12-10T05:06:19.037

2

mat is a double pointer. You need to allocate memory for array of pointers and then allocate memory to each pointers individually.

mat[0] = malloc(sizeof(int) *n ); /* n = Number of elements */

mat[0] should have some memory allocate before writing to it.

Then your scanf() where you have mat[i][j] makes sense. No need to cast malloc()

edited Dec 10 '14 at 05:06

answered Dec 10 '14 at 04:59

Gopi

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@Laura Kent For this answer you can made this in the first loop. – Karthikeyan.R.S Dec 10 '14 at 05:12
Thank you but I'm still a little confused on how to go about implementing your suggestion. Should I use a loop to allocate memory for each of the pointers? – Laura Kent Dec 10 '14 at 05:14

Vlad from Moscow · Answer 2 · 2014-12-10T06:28:38.563

I hope my answer in any case will be useful for you.

If you are using a compiler that supports C99 then you can use at least the following two approaches.

In the first case you can allocate at once a two-dimensional array. In the second case ( that is the case described in your post) you can allocate two one-dimensional arrays that will simulate one two-dimensional array.

Below there is a program that demonstrates the both approaches.

Also do not forget to free allocated arrays. Take into account how the arrays are freed in each case.

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

int main( void )
{
    {
        size_t N = 2;
        size_t M = 3;

        int ( *mat )[M] = malloc( N * M * sizeof( int ) );

        for ( size_t i = 0; i < N; i++ )
        {
            for ( size_t j = 0; j < M; j++ )
            {
                printf( "Input a value for Array[%d][%d]: ", i, j );
                scanf( "%d", &mat[i][j] );
            }
        }

        printf( "\n" );

        for ( size_t i = 0; i < N; i++ )
        {
            for ( size_t j = 0; j < M; j++ )
            {
                printf( "%2d ", mat[i][j] );
            }

            printf( "\n" );
        }

        free( mat );
    }

    {
        size_t N = 2;
        size_t M = 3;

        int **mat = malloc( N * sizeof( int * ) );

        for ( size_t i = 0; i < N; i++ )
        {
            mat[i] = malloc( M * sizeof( M ) );
            for ( size_t j = 0; j < M; j++ )
            {
                printf( "Input a value for Array[%d][%d]: ", i, j );
                scanf( "%d", &mat[i][j] );
            }
        }

        printf( "\n" );

        for ( size_t i = 0; i < N; i++ )
        {
            for ( size_t j = 0; j < M; j++ )
            {
                printf( "%2d ", mat[i][j] );
            }

            printf( "\n" );
        }

        for ( size_t i = 0; i < N; i++ ) free( mat[i] );
        free( mat );
    }

    return 0;
}

In the both cases the program will output the same matrix

1  2  3 
4  5  6

If to eneter these values.

You can test the program at www.ideone.com selecting language strict C99.

this line: int main( void ) is not valid. Rather use: int main() — user3629249, Dec 10 '14 at 21:53
@user3629249 Before C 2011 function main shall be declared as having parameter void. As for my post then I showed program that is written in C89 because even not all modern compilers supports C99. For example till now MS VS does not support C99. So the declaration of main in my post corresponds to C89,C99 Standard. — Vlad from Moscow, Dec 11 '14 at 06:10
@user3629249 That is my example is written using www.ideone.com and selecting language C99 strict. — Vlad from Moscow, Dec 11 '14 at 06:27

2d array initialized with malloc/calloc

2 Answers2