Question

I am trying some programs in c face a problem with this program
can any one tell me what is the problem with this and i also want to know that when in the above case if the pointer value is incremented then will it over write the previous value address as

#include<stdio.h>
int main()
{
    int a=9,*x;
    float b=3.6,*y;
    char c='a',*z;
    printf("the value is %d\n",a);
    printf("the value is %f\n",b);
    printf("the value is %c\n",c);
    x=&a;
    y=&b;
    z=&c;
    printf("%u\n",a);
    printf("%u\n",b);
    printf("%u\n",c);
    x++;
    y++;
    z++;
    printf("%u\n",a);
    printf("%u\n",b);
    printf("%u\n",c);
    return 0;
}

suppose that the value we got in the above program (without the increment in the pointer value )is
65524
65520
65519
and after the increment the value of the pointer is
65526(as 2 increment for the int )
65524(as 4 increment for the float )
65520(as 1 increment for the char variable )

then if in that case will the new pointer address overwrite the content of the previous address and what value be contained at the new address

Answer 1

First of all, I assume your calls to printf are supposed to display the values of x, y and z, or the addresses of a, b, and c, correct? In that case you need to change them to:

printf("%p\n", x);
printf("%p\n", y);
printf("%p\n", z);

These will take into account the size of a pointer on your processor, if that is different from the size of an int. They will also print the address in hexadecimal, which is much more common since addresses can get quite large.

and after the increment the
value of the pointer is 65526(as 2
increment for the int )

Assuming you’re on a platform where int is 2 bytes (on most modern home PCs int is probably 4 now).

then if in that case will the new
pointer address overwrite the content
of the previous address and what value
be contained at the new address
……plz help

You haven’t actually dereferenced any of the pointers here. So the memory occupied by a, b, and c is still untouched; all you’ve done is changed what x, y, and z are pointing to. If you were to dereference one of the pointers after incrementing it, you would be modifying memory past the variable it initially pointed to. So if a is at address 65526 decimal and int is 2 bytes:

int *x = &a; // x points to the integer at "65526" (a)
x++;         // x points to the integer at "65528", a is still at 65526
*x = 5;      // you've just modified the memory at addresses 65528 and 65529.

If 65528 or 65529 contained some other important data (like b or c) you’ve just corrupted your program’s state.