c programming

Question

#include <stdio.h>
void f(char**);
int main()
{
char *argv[] = { "ab", "cd", "ef", "gh", "ij", "kl" };
f(argv);
return 0;
}
void f(char **p)
{
char *t;
t = (p += sizeof(int))[-1];
printf("%s\n", t);
}

can anyone explain this program?

Answer 1

$sizeof(int)$ is compiler dependent

it can be $2$ or $4$ depending on the compiler

if it is $2$, then

 $(p += sizeof(int))[-1]$ can be written as $(p += 2)[-1] $

which can be written as $(p = p+2 - 1)$ which returns address p+1

which is address of $2^{nd}$ element in $argv[]$.

it gives  $cd$

 

if it is $4$, then

 $(p += sizeof(int))[-1]$ can be written as $(p += 4)[-1] $

which can be written as $(p = p+4 - 1)$ which returns address p+3

which is address of $4^{th}$ element in $argv[]$

it will give $gh$

Comments

Is my approach is correct? still i m not getting the clarity...please help

Answer 2

**p contains add of "ab"

According to OperatorPrecedence

1)  sizeof(int) = 4

2)  (p +=sozeof(int))  == (p = p +4)

              and p+4 is address of "ij"

3)  (p += sizeof(int))[-1]  mean "gh"

so t = gh