$i$ = 255, since int is $4B$
which in binary is $0\text{x}11111111$
short int *s = (short int *) &i; Since short is 2B it will point to only $16$ bits.
If it is stored in Little-endian then least significant byte is stored in the lowest address. s will point to initial $2B$ that is$11111111$ $00000000$ = $255$
If it is stored in Big-endian then most significant byte is stored in the lowest address. s will point to $2B$ that is $00000000$ $00000000$ = $0$