C Language, dynamic 3d array - arrays

Question

I have to create a dynamic and 3 dimensional integer-Array in C.
But I have to create the pointers separately and use malloc. I know how to create 2d Array with malloc but I get confused using the following method, and i couldn't really find other question similar to this.
2x2x2 Integer-Array
First step:
int *firstD;
firstD = (int) malloc(2*sizeof(int));
Second step:
int *secondD;
secondD = (int) malloc(2 * firstD * sizeof(int));
Third step:
int *thirdD;
thirdD = (int) malloc(2 * secondD * sizeof(int));
I think maybe I have to add pointers in the starting (int*) and increase it every stepp by one more pointer?

Answer 1

Allocate an array to store pointers to arrays of row pointers.
Allocate arrays to store row pointers.
Allocate arrays to store each rows.
#include <stdlib.h>
int main(void) {
int size1 = 2, size2 = 2, size3 = 2;
int*** array;
array = malloc(sizeof(int**) * size1); // 1
for (int i = 0; i < size1; i++) {
array[i] = malloc(sizeof(int*) * size2); // 2
for (int j = 0; j < size2; j++) {
array[i][j] = malloc(sizeof(int) * size3); // 3
}
}
// free arrays
for (int i = 0; i < size1; i++) {
for (int j = 0; j < size2; j++) {
free(array[i][j]);
}
free(array[i]);
}
free(array);
return 0;
}
I wrote type names explicitly in the above example, but I suggest using dereferencing to obtain size of elements to allocate is better to prevent causing typos.
#include <stdlib.h>
int main(void) {
int size1 = 2, size2 = 2, size3 = 2;
int*** array;
// allocate arrays
array = malloc(sizeof(*array) * size1); // 1
for (int i = 0; i < size1; i++) {
array[i] = malloc(sizeof(*array[i]) * size2); // 2
for (int j = 0; j < size2; j++) {
array[i][j] = malloc(sizeof(*array[i][j]) * size3); // 3
}
}
// free arrays
for (int i = 0; i < size1; i++) {
for (int j = 0; j < size2; j++) {
free(array[i][j]);
}
free(array[i]);
}
free(array);
return 0;
}
Onitted in the above examples to make them simple, but you should check results of malloc() to see if the allocations are successful.