CareerCup

1. The number of iterations depends on sizeof(int) which is platform dependent.
2. The first value will be 255 since the pointer doesn't change if i==0.
3. Then the behavior is undefined.

255
255
255
255

a bit of correction in the question..
actually the question is:
cout<<*(int *)(p + i)<<"\n"; instead of cout<<*(int *)p<<"\n";

if it is (p+i) like the op said...it would give a seg-fault/bus error..otherwise 255 4 times..

what are you guys doing?
simply convert int to char, and back to int.
Then 4 255's.

The data width of the machine will be logically equal to larger than ints. So it is a 32 bit machine, ints should be in 32 bits which is 4 bytes. I think the addressing scheme will be limited only to 32 bits or 4 bytes. so p will point to 4 bytes, p+1 will point to a next memory location which is again 4 bytes (but some random value)

I ran it on my x86 machine. It gave below output -
255
0
0
0

Mine is a little-endian machine. Now, you can guess the answer.

15
15
15
15