passing an array to a function (2d array)

how
is it like


void function (matrix[][])





and calling it

function (matrix[10][10]);
??

try it and see :)

but yes its along the lines of

type arrayname( type [][] )

standard 2d-array function prototype.

More like

Code:

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#define ROWS 10
#define COLS 10

void foo(char a[][COLS])
{
}

int main(void)
{
  char myarray[ROWS][COLS];
  foo(myarray); 
  return 0;
}

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why do u only add the number inside the last brackets only?

Quote:

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Originally posted by revelation437
why do u only add the number inside the last brackets only?

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Because that's the way C works. If you specify one dimension, you have to specify every dimension to the right of it. You always have to specify the last one.

Quzah.

The compiler needs to know the array size in order to do pointer arithmatic. In this snippet:
>>void foochar(char name[]);
the compiler knows that the size of each element in the array is sizeof(char). So when you do name[1], you are adding sizeof(char) to name.

In this example:
>>void foochar(char name[][10]);
the compiler knows the size of name[0][0] is sizeof(char) (as above) and name[0] is size 10*sizeof(char). If you didn't specify the array size in the second brackets, the size equation becomes x*sizeof(char) and your compiler will complain (mine says something like "Size of the type 'signed char[]' is unknown or zero").

Code:

---

#include <stdio.h>

int main(int argc, char *argv[])
{
  char a[10][10][10];
  printf ("sizeof(a[0][0][0]) = %d\n"
          "sizeof(a[0][0])    = %d\n"
          "sizeof(a[0])      = %d\n",
          sizeof(a[0][0][0]),
          sizeof(a[0][0]),
          sizeof(a[0])
          );
  return(0);
}

/*

Output:

sizeof(a[0][0][0]) = 1
sizeof(a[0][0])    = 10
sizeof(a[0])      = 100
 
 */

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