A question about 2D arrays

[HazySmoke)345]

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.

[OneOffDriveByPoster]

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.