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May 30th, 2008

how casting int x[3][2] to int** y ?

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Hello
I have older c functions which I want invoke in new cpp program:

cpp Syntax (Toggle Plain Text)
// Example of old c function
void c_function(int** x, int r, int c)
{ int i, j;
  for (i = 0; i < r; i++)
    {
       for (j=0; j<c;i++) 
           printf("%i ", *(*(x+i)+j));
       printf("\n");
    }
 }
 
// my cpp program, where I would like to invoke c_function
 
int main(int argc, char *argv[])
{
  int  r=3, c=2, 
   x[3][2] =
   { 
      {00, 01}, 
      {10, 11},
      {20, 21}
   };
  c_function(x, r, c);
  return 0;
}
// Example of old c function
void c_function(int** x, int r, int c)
{ int i, j;
  for (i = 0; i < r; i++)
    {
       for (j=0; j<c;i++) 
           printf("%i ", *(*(x+i)+j));
       printf("\n");
    }
 }

// my cpp program, where I would like to invoke c_function

int main(int argc, char *argv[])
{
  int  r=3, c=2, 
   x[3][2] =
   { 
      {00, 01}, 
      {10, 11},
      {20, 21}
   };
  c_function(x, r, c);
  return 0;
}

This does not work. Error in line c_function(x, r, c):
Conversion of first parameter from 'int [3][2]' to 'int **' impossible. (Sorry, no good translation of foreign language message.)

Question: Is it possible to cast 'int [3][2]' to 'int **' ?

I don't like to use ** pointer in cpp program when [][] for two dimensional arrays is possible.

Thanks a lot for your advice.
yap

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yap

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May 30th, 2008

Re: how casting int x[3][2] to int** y ?

The problem is that an array declared as:
int a[ 42 ][ 17 ];
is not the same as
int *a[ 42 ];

The first is a 2D array of 714 (42 x 17) ints.
The second is a 1D array of 42 pointers to int.

The c_function() is looking for something like the second, but you are trying to give it something like the first.

[edit] It is a vagary of the C language that lets you treat pointers like arrays, so however you define it, you can reference a specific int as:
a[ 10 ][ 4 ]

For the first type, it is just a simple multiplication and one dereference to index the element:
*(a + (10 * 42) + 4)

For the second type, it is two dereferences:
*(*(a + 10) + 4)

Hope this helps.

Last edited by Duoas; May 30th, 2008 at 8:02 pm.

Duoas

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Postaholic

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May 31st, 2008

Re: how casting int x[3][2] to int** y ?

How about casting
int* a = (int*)x; ?

Cybulski

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May 31st, 2008

Re: how casting int x[3][2] to int** y ?

Casting won't change the type of thing found found in the array. It will only make the program look at it as if it were a different type -- which in this case is not the correct type.

You can either fix the original function to take the correct type, or you can fix the thing you are passing to it to be the correct type.

BTW, I presume it is a typo at the end of line 6 to i++ instead of j++

Hope this helps.

Duoas

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Jun 1st, 2008

Re: how casting int x[3][2] to int** y ?

yes, its a typo, should have been j.
Now I am convinced that the problem can't be solved with simple cast. What I can do is 1) Changing old c-functions to modern cpp syntax, where the problem with [][const x] still exists. This is also an expensive way. 2) I introduce ** syntax for 2d arrays in cpp program what would allow to use old c functions. Than I need to allocate a vector (first *) and the pointers to further allocated vectors (second *) are stored in first vector. I think this will be the best solution.
Thanks to everybody who has helped me.
yap

yap

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