Latest C Aptitude Question SOLUTION: output? main() { char *p="hai friends",*p1; p1=p; while(*p!='\0') ++*p++; printf("%s %s",p,p1); } Answer. ibj!gsjfoet Explanation: ++*p++ will be parse in the given order _ *p th

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(#M40029243) C question C Keep an EYE

output?
main()
{
char *p="hai friends",*p1;
p1=p;
while(*p!='') ++*p++;
printf("%s %s",p,p1);
}

Answer. ibj!gsjfoet

Explanation: ++*p++ will be parse in the given order
_ *p that is value at the location currently pointed by p will be taken
_ ++*p the retrieved value will be incremented
_ when; is encountered the location will be incremented that is p++ will be executed Hence, in
the while loop initial value pointed by p is ‘h’, which is changed to ‘i’ by executing ++*p and
pointer moves to point, ‘a’ which is similarly changed to ‘b’ and so on. Similarly blank space is
converted to ‘!’. Thus, we obtain value in p becomes “ibj!gsjfoet” and since p reaches ‘’ and p1
points to p thus p1doesnot print anything

Asked In C MAN (10 years ago)

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