#include<stdio.h> main() { char s[]={'a&

#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 Posted / susie

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#include<stdio.h> main() { char s[]={'a','b',..

Answer / susie

Answer :

M

Explanation:

p is pointing to character '\n'.str1 is pointing to
character 'a' ++*p meAnswer:"p is pointing to '\n' and that
is incremented by one." the ASCII value of '\n' is 10. then
it is incremented to 11. the value of ++*p is 11. ++*str1
meAnswer:"str1 is pointing to 'a' that is incremented by 1
and it becomes 'b'. ASCII value of 'b' is 98. both 11 and 98
is added and result is subtracted from 32.

i.e. (11+98-32)=77("M");

Is This Answer Correct ?

15 Yes

1 No

#include<stdio.h> main() { char s[]={'a','b',..

Answer / kiran kumar maharana

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 ?

2 Yes

0 No

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