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Question 
#include<stdio.h>

   main()

   {

      char s[]={'a','b','c','\n','c','\0'};

      char *p,*str,*str1;

      p=&s[3];

      str=p;

      str1=s;

      printf("%d",++*p + ++*str1-32);

   }
 Question Submitted By :: Susie
I also faced this Question!!     Rank Answer Posted By  
 
  Re: #include<stdio.h> main() { char s[]={'a','b','c','\n','c','\0'}; char *p,*str,*str1; p=&s[3]; str=p; str1=s; printf("%d",++*p + ++*str1-32); }
Answer
# 1		 
Answer : 

      77	

      Explanation:

      p is pointing to character '\n'. str1 is pointing to
character 'a' ++*p. "p is pointing to '\n' and that is
incremented by one." the ASCII value of '\n' is 10, which is
then incremented to 11. The value of ++*p is 11. ++*str1,
str1 is pointing to 'a' that is incremented by 1 and it
becomes 'b'. ASCII value of 'b' is 98.

       Now performing (11 + 98 – 32), we get 77("M");

       So we get the output 77 :: "M" (Ascii is 77).
 
Is This Answer Correct ?    0 Yes 0 No
Susie
 

 
 
 
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