Output predict
#include
int main()
{
char *p;
char buf[10] = {1, 2, 3, 4, 5, 6, 9, 8};
p = (buf+1)[5];
printf("%d \n", p);
}
Guess the o/p and what if we change p to *p in last printf statement.
int main()
{
char *p;
char buf[10] = {1, 2, 3, 4, 5, 6, 9, 8};
p = (buf+1)[5];
printf("%d \n", p);
}
Guess the o/p and what if we change p to *p in last printf statement.
p stores address.
ReplyDeletehere p is assigned *(buf+1+5)=9 as its value. The format specifier for pointers is %p but as we are using %d, it will print the signed value, i.e 9.
As we are arbitrarily assigning any address value to p, so if we try to dereference it, it will generate a segmentation fault on unix compilers.
yeah jainendra very good explanation...
ReplyDelete