Traverse through layers of array using pointer to layer of array

Question

                         |--------|     
//                     / |4  4  4 |     
//                   |--------| 4 | 
//                 / |3  3  3 | 4 | 
//               |---------|3 |   |
//             / | 2  2  2 |3 | /
//            |---------|2 |__|
//            | 1  1  1 |2 | /
//            | 1  1  1 |__| 
//            | 1  1  1 | /
//            |_________|
double arr[4][3][3] = {{1,1,1,1,1,1,1,1,1},{2,2,2,2,2,2,2,2,2},{3,3,3,3,3,3,3,3,3},{4,4,4,4,4,4,4,4,4}};

I consider that this array consists of 4 layers.

I want to create pointer to layer of array and traverse through layers of that array using pointer.

I try :

double (*pp1)[sizeof(arr[0]) / sizeof(ar[0][0][0])]; 
pp1 = arr[0]; 

and get error from intelIsense: value of type (double (*)(3) can`t be assigned to double(*)(9)

Answer 1

So if you do:

int i;
double arr[4][3][3] = {{1,1,1,1,1,1,1,1,1},{2,2,2,2,2,2,2,2,2},
                       {3,3,3,3,3,3,3,3,3},{4,4,4,4,4,4,4,4,4}};

double (*pp3)[3][3];    
pp3 = arr;    
for (i = 0; i <= 3; i++)    
{   printf("pp3 + %d is %f \n", i, ***(pp3 + i));
}   

Then you get the desired behavior. The problem with your code is that, just as the compiler is telling you, you are trying to assign a pointer to an array of 9 double (double (*pp1)[sizeof(arr[0] / sizeof(arr[0][0][0])] evaluates to double (*pp1)[9] ), but you need a pointer to an array of 3 of array of 3, which is what you declared with double arr[4][3][3]. From my tests, gcc will accept the double (*pp1)[9] with a compiler warning, which is what I tried to get at in my comment below. Hope that clears things up.

If you want to keep this general, then what you really want is double (*pp3)[sizeof(arr[0]) / sizeof(arr[0][0])][sizeof(arr[0][0]) / sizeof(arr[0][0][0])], which is a bloody nightmare.

EDIT: Forgot a dereference... Should've copy/pasted haha. Also added explanation about the question's code behavior. Fixed as per comments.