Pointer Arithmetic for a 3D Array

Question

I know how pointers work for a 1D and 2D array, but I'm trying to figure it out for a 3D array.

For a 1D array:

V[i] ==> *(V+i)

for a 2D array with C columns

M[i][j] ==> *(M+(C*i+j))

What would be the pointer arithmetic for a 3D array?

user529758 · Accepted Answer · 2013-04-22 22:38:49Z

3

Rewriting the subscription using pointer syntax then expanding the pointer arithmetic gives:

M[i][j][k] = *(*(*(M + i) + j) + k) = *(&M[0][0][0] + x * y * i + y * j + k)

where x, y and z (the latter one unused) are the dimensions of the array.

answered Apr 22, 2013 at 22:38

user529758

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newacct · Accepted Answer · 2013-04-23 09:03:49Z

3

M[i][j] ==> *(M+(C*i+j))

wrong. It is

M[i][j] ==> *(*(M+i)+j)

for 3D

M[i][j][k] ==>  *(*(*(M+i)+j)+k)

edited Apr 23, 2013 at 9:03

newacct

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answered Apr 22, 2013 at 22:33

yngccc

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4 Comments

andrjas Over a year ago

you're doing triple dereferencing here?

Barmar Over a year ago

Isn't this for an array of pointers, not an n-dimensional array?

user529758 Over a year ago

@yngum No. 2D arrays aren't arrays of pointers. 3D arrays aren't arrays of arrays of pointers.

Barmar Over a year ago

It works because of the way arrays decay to pointers in various contexts in C. But it's not very helpful for understanding how multiple dimensions map to row-major indexing in memory, which seems to be what the OP is looking for.

Barmar · Accepted Answer · 2013-04-22 22:34:06Z

2

If the 3D array has C columns and R rows, each slice is C*R elements.

M[i][j][k] = *(M + C*R*i + C*j + k)

answered Apr 22, 2013 at 22:34

Barmar

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Comments

andrjas · Accepted Answer · 2013-04-22 22:35:10Z

0

for a matrix M[X][Y][Z]

M[i][j][k] = *(M+(Y*Z*i+Z*j+k))

answered Apr 22, 2013 at 22:35

andrjas

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