C Programming - Pointers - Discussion

printf("%s\n", s++ +3);

In the above statement, ++ after s is post-increment, s represents the address of p and s+3 will give the address of c. Here, s++ +3 is the same as s+3 because ++ is post-increment.

int main()
{
printf("%x\n", 9["IndiaSAIRAM"]);
return 0;
}

What will be the output and please explain how?

@ALL.

#include<stdio.h>
int main()
{
Step1: char str[] = "peace";
/* It will declare str[] array as character and store peace in it.*/

Step 2: char *s = str;
/* It will declare a pointer variable (s) as character and store the content of str. */

Step3: printf("%s\n", s++ +3);
/* It will print the string (as specified %s) , the next question is what?

(i) s++ --> It will point to "p" because it is post_increment operator and still contain string as "peace".
(ii) s++ +3 = add three positions (i.e. At present , it is pointing to s[0] but after incrementing s[0+3] =s[3] , and it is pointing to "c" position no. 3) and prints further (i.e. "ce").
return 0;
}

@All.

Please consider this.

#include<stdio.h>

int main()
{
char str[]="peace";
char *s = str;
printf("%s\n",++s +3);
return 0;
}

++ is the unary operator and + is a binary operator.
unary given higher precedence over binary.
so ++ should consider.

Is it right?

Thanks @Monika.

Thanks @Monika.

=> s++, executes later after printing the string for the first time.

Rest what remains is s+3, which makes the pointer point to c and from there it will print the remaining string.

int main()
{
char str[] = "peace";
char *s = str;
printf("%s\n", ++s +2);
return 0;
}

It will print as "ce". Because pre-increment and post-increment concept is also included in this code.

Printf ("%c\n", s++ +3);// This will print only c right?