Answer / susie

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 ‘\0’
and p1 points to p thus p1doesnot print anything.

Answer / sourav punoriyar

checked in gcc.

it gives segmentation fault(core dump),in gcc...
because the char *p="hai friends",is a pointer pointing to
this string in the code section,(this string is present in
code section.)
now,
++*p=this is ++(*p)=h+1=i,and stores it in p,but data in
code section cannot be modified so core dump.
if
*p++,first dereference and then increases the pointer....so
it will point to a now.

Answer / sourav punoriyar

but in turbo c it can be the given ans ,as given by susie,
as there it gets stored in datasection which is modifiable

main ( ) { static char *s[ ] = {“black”, “white”, “yellow”, “violet”}; char **ptr[ ] = {s+3, s+2, s+1, s}, ***p; p = ptr; **++p; printf(“%s”,*--*++p + 3); }

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