Computer Science Canada

A question about 2D arrays
Author:  HazySmoke)345 [ Sun Oct 31, 2010 10:12 am ]
Post subject:  A question about 2D arrays
So, to the best of my understanding, there are two TYPES of 2D arrays.

code:
// Type 1
int a[3][4];

// Type 2
int **b = (int **) malloc (3 * sizeof(int *));
for (i = 0; i < 3; ++i)
    b[i] = (int *) malloc (4 * sizeof(int));

//How to access array elements
a[1][1] = 74;
b[1][1] = 74;


In this case, a is an array of array, all of the 12 elements are stored in a continous chunk. But b is an array of pointers, all the rows are stored in different places. If I want, I can even make the rows to have different lengths. We access the 2D arrays the same way, but how they are laid out in memory is a little different.

Here's my first question. Can I create the Type 1 2D array dynamically, using malloc? I think the answer is yes, but someone please confirm that I'm doing this right:

code:
int (*c)[4] = (int (*)[4]) malloc (3 * sizeof(int [4]));

c[1][1] = 74;


Also, why doesn't this give compile errors?

code:
#include <stdlib.h>
int foo(int (*a)[]) {}
int main(){
    int a[3][4];
    int (*c)[4] = (int (*)[4]) malloc (3 * sizeof(int [4]));
       
    foo(a); //<-- HERE
    foo(c); //<-- AND HERE
       
    free(c);
}


Note that I did not specify the column width in foo, You'd think it should be like this

code:
int foo(int (*a)[4]) {}


And lastly, what's the real difference between MALLOC and CALLOC? I get the impression that we *should* use malloc for creating single objects and calloc for arrarys, but time and time again I see people creating arrays using malloc.
Author:  OneOffDriveByPoster [ Wed Nov 03, 2010 4:41 pm ]
Post subject:  Re: A question about 2D arrays
code:
int (*c)[4] = (int (*)[4]) malloc (3 * sizeof(int [4]));

c[1][1] = 74;
Yes, that is okay (assuming malloc did not return null). malloc() does not initialize the memory it returns, calloc() sets all bytes to zero.

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