Why does pointer differencing not work in this case?

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Today I am trying to learn better pointer differencing and I edited some code from a C-book. The problem is that where the book says it should be 1 my result is -0.100000. Here is the code:

// Jones, Bradley L.; Peter Aitken; Dean Miller. C Programming in One Hour a Day, Sams Teach Yourself (pp. 339-340). Pearson Education. Kindle Edition.
   /* ptr_math.c--Demonstrates using pointer arithmetic to
      access array elements with pointer notation. */

   #include <stdio.h>
   #define MAX 10

   // Declare and initialize an integer array.

   int i_array[MAX] = { 0,1,2,3,4,5,6,7,8,9 };

  // Declare a pointer to int and an int variable.

  int *i_ptr, *i2_ptr, *i3_ptr, count;

  // Declare and initialize a float array.

  float f_array[MAX] = { .0, .1, .2, .3, .4, .5, .6, .7, .8, .9 };

  // Declare a pointer to float.

  float *f_ptr, f2_ptr, f3_ptr;

  int main( void )
  {
      /* Initialize the pointers. */

      i_ptr = i_array;
      i2_ptr = i_array[2];
      i3_ptr = i_array[3];
      f_ptr = f_array;
      f2_ptr = f_array[2];
      f3_ptr = f_array[3];

      /* Print the array elements. */

    //  for (count = 0; count < MAX; count++)
    //      printf("%d\t%f\n", *i_ptr++, *f_ptr++);
    for (count = 0; count < MAX; count++)
    {
    printf("%d\t%f\n", *i_ptr, *f_ptr );
    i_ptr++;
    f_ptr++;
    }
    printf("\n%d\n", i2_ptr - i3_ptr );
    printf("\n%f\n", f2_ptr - f3_ptr );

      return 0;
  }

So it's the last 2 printf's where I difference with pointers. I understand that they are 1 position away from each other but the float value is not 1 but -0.100000. Why?

What I have tried:

I don't remember today. It has passed some time since I edited the code from the C-book.

Posted 6-Aug-21 4:11am

Mieczyslaw1683

Updated 6-Aug-21 6:31am

v2

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jeron1 6-Aug-21 10:43am

try something like,

i2_ptr = &i_array[2];

Mieczyslaw1683 6-Aug-21 10:56am

Hmm, I don't understand. The result is weird with that & symbol before. However, this is what I understand how the code should look and work correctly:

// Jones, Bradley L.; Peter Aitken; Dean Miller. C Programming in One Hour a Day, Sams Teach Yourself (pp. 339-340). Pearson Education. Kindle Edition.
   /* ptr_math.c--Demonstrates using pointer arithmetic to
      access array elements with pointer notation. */

   #include <stdio.h>
   #define MAX 10

   // Declare and initialize an integer array.

   int i_array[MAX] = { 0,1,2,3,4,5,6,7,8,9 };

  // Declare a pointer to int and an int variable.

  int *i_ptr, *i2_ptr, *i3_ptr, count;

  // Declare and initialize a float array.

  float f_array[MAX] = { .0, .1, .2, .3, .4, .5, .6, .7, .8, .9 };

  // Declare a pointer to float.

  float *f_ptr, *f2_ptr, *f3_ptr;

  int main( void )
  {
      /* Initialize the pointers. */

      i_ptr = i_array;
      i2_ptr = i_array[2];
      i3_ptr = i_array[3];
      f_ptr = f_array;
      f2_ptr = f_array[2];
      f3_ptr = f_array[3];

      /* Print the array elements. */

    //  for (count = 0; count < MAX; count++)
    //      printf("%d\t%f\n", *i_ptr++, *f_ptr++);
    for (count = 0; count < MAX; count++)
    {
    printf("%d\t%f\n", *i_ptr, *f_ptr );
    i_ptr++;
    f_ptr++;
    }
    printf("\n%d\n", i2_ptr - i3_ptr );
    printf("\n%f\n", f2_ptr - f3_ptr );

      return 0;
  }

The f2_ptr and f3_ptr are real pointers but the error when trying this code is:
"... \Giraffe\main.c|34|error: incompatible types when assigning to type 'float *' from type 'float'|
... \Giraffe\main.c|35|error: incompatible types when assigning to type 'float *' from type 'float'|"

jeron1 6-Aug-21 11:02am

Like Griff said,

i2_ptr = i_array[2];

You've specified i2_ptr as a pointer, i_array[2] is not a pointer but the float value at index 2 of the array. When you have,

i2_ptr - i3_ptr

what is your expectation?

I didn't read the whole thing prior to commenting the first time, I apologize.

Richard MacCutchan 6-Aug-21 11:30am

f2_ptr = f_array[2];

You cannot assign a floating point value to a pointer. If you want it to point to the third element of the array then you need to add the addressof operator, thus:

f2_ptr = &f_array[2];

Richard MacCutchan 6-Aug-21 11:32am

If you really want to learn C then start with The C Programming Language - Wikipedia[^], the best book ever written on the subject.

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Mieczyslaw1683 · Answer 1 · 2021-08-06T06:31:00

Now I have made correct pointer arithmetic subtractions (in 2 types of arrays), I think. The code is below:

// Jones, Bradley L.; Peter Aitken; Dean Miller. C Programming in One Hour a Day, Sams Teach Yourself (pp. 339-340). Pearson Education. Kindle Edition.
   /* ptr_math.c--Demonstrates using pointer arithmetic to
      access array elements with pointer notation. */

   #include <stdio.h>
   #define MAX 10

   // Declare and initialize an integer array.

   int i_array[MAX] = { 0,1,2,3,4,5,6,7,8,9 };

  // Declare a pointer to int and an int variable.

  int *i_ptr, *i2_ptr, *i3_ptr, count;

  // Declare and initialize a float array.

  float f_array[MAX] = { .0, .1, .2, .3, .4, .5, .6, .7, .8, .9 };

  // Declare a pointer to float.

  float *f_ptr, *f2_ptr, *f3_ptr;

  int main( void )
  {
      /* Initialize the pointers. */

      i_ptr = i_array;
      i2_ptr = &i_array[2];
      i3_ptr = &i_array[3];
      f_ptr = f_array;
      f2_ptr = &f_array[2];
      f3_ptr = &f_array[3];

      /* Print the array elements. */

    //  for (count = 0; count < MAX; count++)
    //      printf("%d\t%f\n", *i_ptr++, *f_ptr++);
    for (count = 0; count < MAX; count++)
    {
    printf("%d\t%f\n", *i_ptr, *f_ptr );
    i_ptr++;
    f_ptr++;
    }
    printf("\n%d\n", i2_ptr - i3_ptr );
    printf("\n%d\n", f2_ptr - f3_ptr );

      return 0;
  }

Mieczyslaw1683 · Answer 2 · 2021-08-06T05:56:00

Okay. I have learned to do some pointer arithmetic and I have seen how far apart 2 elements in two different types of arrays are from each other. Here is the code:

// Jones, Bradley L.; Peter Aitken; Dean Miller. C Programming in One Hour a Day, Sams Teach Yourself (pp. 339-340). Pearson Education. Kindle Edition.
   /* ptr_math.c--Demonstrates using pointer arithmetic to
      access array elements with pointer notation. */

   #include <stdio.h>
   #define MAX 10

   // Declare and initialize an integer array.

   int i_array[MAX] = { 0,1,2,3,4,5,6,7,8,9 };

  // Declare a pointer to int and an int variable.

  int *i_ptr, *i2_ptr, *i3_ptr, count;

  // Declare and initialize a float array.

  float f_array[MAX] = { .0, .1, .2, .3, .4, .5, .6, .7, .8, .9 };

  // Declare a pointer to float.

  float *f_ptr, *f2_ptr, *f3_ptr;

  int main( void )
  {
      /* Initialize the pointers. */

      i_ptr = i_array;
      i2_ptr = i_array[2];
      i3_ptr = i_array[3];
      f_ptr = f_array;
      f2_ptr = &f_array[2];
      f3_ptr = &f_array[3];

      /* Print the array elements. */

    //  for (count = 0; count < MAX; count++)
    //      printf("%d\t%f\n", *i_ptr++, *f_ptr++);
    for (count = 0; count < MAX; count++)
    {
    printf("%d\t%f\n", *i_ptr, *f_ptr );
    i_ptr++;
    f_ptr++;
    }
    printf("\n%d\n", i2_ptr - i3_ptr );
    printf("\n%d\n", f2_ptr - f3_ptr );

      return 0;
  }

For the future, I would like to understand why integer pointers don't need the "&" symbol before array names (with elements marked)?

P.s. If it would be the same for both the int array and float array it would look more logical.

Mieczyslaw1683 · Answer 3 · 2021-08-06T05:24:00

Okay. So I have learned that pointers with adresses (adress-pointers) need the & before the, for example, array name. Then they can be int's without problems because it's just a value. So now I got it to work. Here is the code:

// Jones, Bradley L.; Peter Aitken; Dean Miller. C Programming in One Hour a Day, Sams Teach Yourself (pp. 339-340). Pearson Education. Kindle Edition.
   /* ptr_math.c--Demonstrates using pointer arithmetic to
      access array elements with pointer notation. */

   #include <stdio.h>
   #define MAX 10

   // Declare and initialize an integer array.

   int i_array[MAX] = { 0,1,2,3,4,5,6,7,8,9 };

  // Declare a pointer to int and an int variable.

  int *i_ptr, *i2_ptr, *i3_ptr, *i4_ptr, *i5_ptr, count;

  // Declare and initialize a float array.

  float f_array[MAX] = { .0, .1, .2, .3, .4, .5, .6, .7, .8, .9 };

  // Declare a pointer to float.

  float *f_ptr;

  int main( void )
  {
      /* Initialize the pointers. */

      i_ptr = i_array;
      i2_ptr = i_array[2];
      i3_ptr = i_array[3];
      f_ptr = f_array;
      i4_ptr = &f_array[2];
      i5_ptr = &f_array[3];

      /* Print the array elements. */

    //  for (count = 0; count < MAX; count++)
    //      printf("%d\t%f\n", *i_ptr++, *f_ptr++);
    for (count = 0; count < MAX; count++)
    {
    printf("%d\t%f\n", *i_ptr, *f_ptr );
    i_ptr++;
    f_ptr++;
    }
    printf("\n%d\n", i2_ptr - i3_ptr );
    printf("\n%d\n", i4_ptr - i5_ptr );

      return 0;
  }

OriginalGriff · Answer 4 · 2021-08-06T04:57:00

Solution 1

Because you aren't working with pointers:

C

float *f_ptr, f2_ptr, f3_ptr;
    f_ptr = f_array;
    f2_ptr = f_array[2];
    f3_ptr = f_array[3];

f_ptr is a pointer, but f2_ptr and f3_ptr aren't = they are float values, you you copy the content of two array elements into them before you do the subtraction: 0.2 and 0.3
Subtract 0.3 from 0.2 and you get ... -0.1

i2_ptr and i3_ptr are pointers, so when you subtract them, you get the number of elements between them.

Posted 6-Aug-21 4:57am

OriginalGriff

Comments

Mieczyslaw1683 6-Aug-21 10:59am

Ok. So yes. I have noticed that I skipped the "*" symbol before so they became float values. Because I got this error when making pointers to the f_array elements: "... \Giraffe\main.c|34|error: incompatible types when assigning to type 'float *' from type 'float'|
... \Giraffe\main.c|35|error: incompatible types when assigning to type 'float *' from type 'float'|"

Why do I get that errors when making pointers to the array elements?

Why does pointer differencing not work in this case?

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