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main()
{
char *p="hai friends",*p1;
p1=p;
while(*p!='\0') ++*p++;
printf("%s %s",p,p1);
}
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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.
| Is This Answer Correct ? | 6 Yes | 1 No |
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.
| Is This Answer Correct ? | 2 Yes | 0 No |
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
| Is This Answer Correct ? | 1 Yes | 0 No |
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