convert array to two dimensional array by pointer

Question

Is it possible to convert a single dimensional array into a two dimensional array?

i first tought that will be very easy, just set the pointer of the 2D array to the beginning of the 1D array like this:

int foo[] = {1,2,3,4,5,6};
int bla[2][3] = foo;

because i can easily create an two dimensional array like this:

int bla[2][3] = {1,2,3,4,5,6};

so the question is now, is there a way to convert it via the pointer?

Answer 1

You can't initialise an int bla[2][3] with an int* (what foo becomes in that context).

You can achieve that effect by declaring a pointer to arrays of int,

int (*bla)[3] = (int (*)[3])&foo[0];

but be sure that the dimensions actually match, or havoc will ensue.

Answer 2

I know you specificed pointers... but it looks like you're just trying to have the data from an array stored in a 2D array. So how about just memcpy() the contents from the 1 dimensional array to the two dimensional array?

int i, j;
int foo[] = {1, 2, 3, 4, 5, 6};
int bla[2][3];
memcpy(bla, foo, 6 * sizeof(int));
for(i=0; i<2; i++)
   for(j=0; j<3; j++)
      printf("bla[%d][%d] = %d\n",i,j,bla[i][j]);

yields:

bla[0][0] = 1
bla[0][1] = 2
bla[0][2] = 3
bla[1][0] = 4
bla[1][1] = 5
bla[1][2] = 6

That's all your going for, right?

Answer 3

int (*blah)[3] = (int (*)[3]) foo; // cast is required

for (i = 0; i < 2; i++)
  for (j = 0; j < 3; j++)
    printf("blah[%d][%d] = %d\n", i, j, blah[i][j]);

Note that this doesn't convert foo from a 1D to a 2D array; this just allows you to access the contents of foo as though it were a 2D array.

So why does this work?

First of all, remember that a subscript expression a[i] is interpreted as *(a + i); we find the address of the i'th element after a and dereference the result. So blah[i] is equivalent to *(blah + i); we find the address of the i'th 3-element array of int following blah and dereference the result, so the type of blah[i] is int [3].