How to allocate and declare a 3D of array of structs in C?

Question

How do you allocate and declare a 3D array of structs in C? Do you first allocate the array or declare it? I feel like you have to allocate it first so you can declare it so it is on the heap, but then how do you allocate something that hasn't been made yet? Also, should you allocate it all at once or element by element? Also am i putting the structs into the array correctly? My guess on how to do it would be:

header.h

struct myStruct{
   int a;
   int b;
}; 
typedef struct myStruct myStruct_t;

main.c

#include "header.h"
#include <stdio.h>
#include <stdlib.h>
int main(void){

    int length=2;
    int height=3;
    int width =4;
    myStruct_t *elements;
    struct myStruct arr = (*myStruct_t) calloc(length*height*width, sizeof(myStruct);
    //zero based array
    arr[length-1][height-1][width-1];

    int x=0;
    while(x<length){
        int y=0;
        while(y<height){
            int z=0;
            while(z<depth){
                arr[x][y][z].a=rand();
                arr[x][y][z].b=rand();
                z++;
            }
            y++;
        }
        x++;
    }
    return 0;
}    

Answer 1

The easy way is:

 myStruct_t (*arr2)[height][width] = calloc( length * sizeof *arr );

Then your loop can access arr2[x][y][z].a = rand(); and so on. If you're not familiar with this way of calling calloc, see here. As usual with malloc, check arr2 against NULL before proceeding.

The triple-pointer approach is not really a practical solution. If your compiler does not support variably-modified types then the array should be flattened to 1-D.