What is the purpose of the gate2011 Program in the C programming language?

[sam topper.]

#include<stdio.h>
void main()
{
char c[]="gate2011";
char *p=c;
printf("%s",p+p[3]-p[1]);
}

 

[CompuChip]

Doesn't look like rocket science. What do you think will happen? What actually happens when you run it?

 

[jk22]

It apparently takes the address of the string + 65-61= adr + 4. Now I don't know but this seems not aligned, but it means maybe adr + 4 bytes, hence it maybe prints '2011', but I have not checked this for real.

 

[CompuChip]

The trick is the way that pointers and arrays are equivalent in C++.
If you have an array

int[] a;

then a is a pointer to an int, and &a[4] and *(a + 4) both point to address of the 5th array element.
So the pointer p points at the beginning of the string, p[4] - p[1] is the length of the memory block used by the first four characters and the sum is a pointer to the place in memory where the '2' is stored. Since strings are terminated by a null (\0) the statement is equivalent to

    char p[] = { '2', '0', '1', '1', '\0' };
    printf("%s", p);

You could actually reach the same thing through

#include<stdio.h>
void main()
{
    char c[]="gate2011";
    // No need to introduce a new name, c is already a char*
    // char* p = c; 
    printf("%s", (c + 4));
}