how does following program work pointer manipulation [duplicate]

Question

Possible Duplicate:
A c program from GATE paper

Here is a program which is working

#include<stdio.h>
int main ()
{
char c[]="GATE2011";
char *p=c;
printf("%s",p+p[3]-p[1]);
}

The output is

2011

Now comes the problem I am not able to understand the operation p+p[3]-p[1] what is that pointing to?

My understanding is when I declare some thing like

char c[]="GATE2011"

Then c is a pointer pointing to a string constant and the string begins with G. In next line *p=c; the pointer p is pointing to same address which c is pointing. So how does the above arithmetic work?

Answer 1

p[3] is 'E', p[1] is 'A'. The difference between the ASCII codes A and E is 4, so p+p[3]-p[1] is equivalent to p+4, which in turn is equivalent to &p[4], and so points to the '2' character in the char array.

Anyone found writing this sort of thing in production code would be shot, though.

Answer 2

It's pretty horrid code. (p+p[3]-p[1]) is simply adding and subtracting offsets to p. p[3] is (char)'E', which is 69 in ASCII. p[1] is (char)'A', which is 65 in ASCII. So the code is equivalent to:

(p+69-65)

which is:

(p+4)

So it's simply offseting the pointer by 4 elements, before passing it to printf.

Technically, this is undefined behaviour. The first part of that expression (p+69) offsets the pointer beyond the end of the array, which is not allowed by the C standard.