pointers

[srinurocks]

hello guys,

here is my problem


if wer are adding a value to address like

if the address location is 1002

if we add 1002+1 what it will show

if a[0]'s current address position is 1002

main()
{
int a[2]={0,9};
printf("%u",a[0]+1);

}

what is the o/p

bye
srinu

[shaik786]

>if we add 1002+1 what it will show
It depends on the type of data this location is holding. Assuming the following:
sizeof(char) = 1 (1002 + 1 = 1003)
sizeof(int) = 4 (1002 + 1 = 1006)
sizeof(long) = 8 (1002 + 1 = 1010)

>main()
>{
>int a[2]={0,9};
>printf("%u",a[0]+1);
>}
Output will be: 1
That's because you are adding the contents at that location. Try (a + 1) instead.

[Fordy]

Try it yourself....

We arent here to do your homework

[Monster]

Some tips:
1. use int main() and return something
2. we don't do homework
3. use code tags:

&#91CODE&#93 /* your code */ &#91/CODE&#93

[Prelude]

Assuming you were going for something more like this:

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

int main ( void )
{
  int a[2] = {0,9};
  int *b = a;
  printf ( "%p\n%p\n", b, b + 1 );
  return EXIT_SUCCESS;
}

The output would be platform-dependent because the actual size of data types isn't specified by the C standard. On my system, the output of this program gave me

0012FF78
0012FF7C

Which is consistent with an integer being 4 bytes on my computer.

-Prelude