Program on pointers

thelink123 · 07-21-2009, 10:39 PM

Hi

Code:

   main ()
   {
      int a[] = {10,60,30,40,50};
      char *p;
      p = (char *)a;
      printf("%d",*((int *)p+4));
   }

The ouput of this code is 50. I know why this output is obtained. Now take a look at the next piece of code

main ()
{
int a[] = {10,60,30,40,50};
char *p;
p = (char *)a;
printf("%d",*((int *)(p+4)));
}

The output of this code is 60. I need the explanation for this output. I know this is something to do with type-casting. Kindly explain the output in terms of the number of bytes moved on adding 4 to reach the element '60' in the array a[].

Regards,
TheLink

Sebastiani · 07-21-2009, 10:48 PM

>> I need the explanation for this output. I know this is something to do with type-casting.

When you add an offset to a pointer it advances a number of bytes equal to the the offset multiplied by the size of the type. So if 'p' is a char* then p + 4 == ((char*)p) + (4 * sizeof(char)) == 4 bytes (on most systems) in advance, whereas if 'p' is an int* then p + 4 == ((char*)p) + (4 * sizeof(int)) == 16 bytes (on most systems) in advance.

07-21-2009, 10:39 PM	#1
thelink123 Registered User Join Date: Jul 2009 Location: Kerala, India Posts: 3	Program on pointers - explanation needed Hi Code: main () { int a[] = {10,60,30,40,50}; char p; p = (char )a; printf("%d",((int )p+4)); } The ouput of this code is 50. I know why this output is obtained. Now take a look at the next piece of code main () { int a[] = {10,60,30,40,50}; char p; p = (char )a; printf("%d",((int )(p+4))); } The output of this code is 60. I need the explanation for this output. I know this is something to do with type-casting. Kindly explain the output in terms of the number of bytes moved on adding 4 to reach the element '60' in the array a[]. Regards, TheLink
thelink123 is offline

07-21-2009, 10:48 PM	#2
Sebastiani Guest Join Date: Aug 2001 Posts: 4,923	>> I need the explanation for this output. I know this is something to do with type-casting. When you add an offset to a pointer it advances a number of bytes equal to the the offset multiplied by the size of the type. So if 'p' is a char* then p + 4 == ((char)p) + (4 sizeof(char)) == 4 bytes (on most systems) in advance, whereas if 'p' is an int* then p + 4 == ((char)p) + (4 sizeof(int)) == 16 bytes (on most systems) in advance.
Sebastiani is offline

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