Counting the pointers in a pointer and an invalid read of size 8...

**MutantJohn** · 07-29-2013

Hello All,

I'm trying to count the number of pointers in a pointer.

Basically, if we have

Code:

int **x;

then I wanna know how many *x's are in **x.

I keep getting an invalid read error from valgrind every time I run this code even though it seems to be counting properly.

Code:

#include <stdio.h>
#include <stdlib.h>


int main(void) {


   int **x = malloc(4*sizeof(*x));


   for (int i=0; i<4; i++) {


      x[i] = malloc(4*sizeof(*x[i]));
      *x[i] = 0;
   }


   int i=0;


   while(x[i]) {


      if (x[i]) i++;
   }


   printf("Total number of pointers, %d\n", i);


   for (int i=0; i<4; i++)
      free(x[i]);


   free(x);


   return 0;
}

**whiteflags** · 07-29-2013

One problem I have with this code is that it's not clear when this while loop ends.

Code:

   while(x[i]) {
      if (x[i]) i++;
   }

The way things are, the loop needs to end on NULL, or some other sentinel value. I could set it up like so:

Code:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main(void)
{
  const int xDim = 4;
  const int yDim = xDim;
  int i;
  int **x;
  
  x = malloc((xDim + 1) * sizeof x[0]);
  for (i = 0; i < xDim; i++) {
    x[i] = malloc(yDim * sizeof *x[0]);
    memset(x[i], 0, sizeof *x[0] * yDim);
  }
  x[xDim] = NULL;

  
  for (i = 0; x[i]; i++) ;

  printf("%d pointers in x.\n", i);
  
  for (i = 0; x[i]; i++) free(x[i]);
  free(x);
  
  return 0;
}

C code - 29 lines - codepad

**MutantJohn** · 07-29-2013

Ah, I see now. I was hoping that I wouldn't have to construct an extra memory value just to set it to NULL. Oh well.

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