Complete the below function which dynamically allocates space to a 3d array of doubles, initializes all values to 0, and returns a pointer to the space.

1. double ***alloc3dArrayOfInts( int length, int width, int depth) {
2. double ***array3d = malloc( * sizeof(double **) );
3. for(int i=0; i< length ;i++) {
4. = malloc(width * sizeof(double *) );
5. for(int j=0; j< ;j++) {
6. = malloc(depth * sizeof(double) );
7. }
8. }
9. return array3d;
10. }

Explanation:

1. double ***alloc3dArrayOfInts( int length, int width, int depth) {

2. double ***array3d = malloc(length * sizeof(double **) );

3. for(int i=0; i< length ;i++) {

4. array3d[i] = malloc(width * sizeof(double *) );

5. for(int j=0; j< width;j++) {

6. array3d[i][j] = malloc(depth * sizeof(double) );

7. }

8. }

9. return array3d;

10. }

answer; ron will have to compare ( b3: b4); /// good luck

the answer to your question is ⇔ d. the nether

explanation: