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

Answer 1

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 2

Yes, if you can use an array of pointers:

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

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].

Answer 4

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 5

You could use union to alias one array into the other:

#include <stdio.h>

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

int main(void)
{
  int i, j;

  for (i = 0; i < 6; i++)
    printf("u.foo[%d]=%d ", i, u.foo[i]);
  printf("\n");

  for (j = 0; j < 2; j++)
  {
    for (i = 0; i < 3; i++)
      printf("u.bla[%d][%d]=%d ", j, i, u.bla[j][i]);
    printf("\n");
  }

  return 0;
}

Output (ideone):

u.foo[0]=1 u.foo[1]=2 u.foo[2]=3 u.foo[3]=4 u.foo[4]=5 u.foo[5]=6 
u.bla[0][0]=1 u.bla[0][1]=2 u.bla[0][2]=3 
u.bla[1][0]=4 u.bla[1][1]=5 u.bla[1][2]=6