UGC NET CSE | June 2010 | Part 2 | Question: 12

Question

What will be the output of the following c-code?

void main () 
{ 
   char *p= "ayqm";
   char c;  
   c=++*p;  
   printf ("%c" , c);  
}

• A. $a$
• B. $c$ 
• C. $b$
• D. $q$

Comments

when you can this code    

void main ( )
{
char *P = "ayqm" ;
char c;
c =++ *P  ;
printf ("%c", c);
}

you will get Segmentation error but if you write (*P+1) then you get b as anwer

Answer 1

Option C

++*p   ====>++(*p) due to precedence table in C

(*p)=a

++(*p)=b; but here *p is a string constant which cannot be changed

so there is runtime error

there is one more error

in intialization *P capital Pis used and

in line c=++(*p); small p is used which is not declared

Comments

Don,t you think here ++*p will cause problem(i.e. runtime issue) ,Since we are trying to change content of string constant which is not possible(bcz they are stored in ROM).

---

p is a pointer pointing to address of a

Now value of p is incrementing.

So, why not ASCII value of a will not increment?

---

Ok.

What is the meaning expression ++*p.

Here the priority goes to * first.So the expression increments the value pointed by p and update it.

Which will like this *p=*p+1.So the expression trying to change the content of string constant .But "ayqm" is a string constant which is stored in read only location hence can,t be modified.Attempting to modify the string literal is undefined behavior:Thus runtime error.

If not clear try to run this program and see the difference:

#include<stdio.h>
 void main ()
{
   char *p= "ayqm";
   char c;
   c=*p+1;
   printf ("%c" , c);
}