In their study guide , Roberts ,Heller and Ernest say "shifts of ints use only the low-order 5 bits ,and shifts of longs use only the low-order 6bits , of the right operand".What does this mean?Example? jeff

Suppose you want to shift int X to the right over B bits, then you would write simply: X >> B. In that case only the 5 lowest-order bits of B are used, which basically means that you can shift over maximum 32 bits. When you want to shift a long, the 6 lowest-order bits of B are used, so you can shift over maximum 64 bits. If you try to shift over more than 32 (long: 64) bits, the highest-order bits are ignored, this is the same as doing modulo 32 (long: 64). So shifting over 35 bits results in the same as shifting over 3 bits. Some examples: int X; X >> 5; // shifts 5 bits to the right X >> 32; // shifts 0 bits to the right, does nothing X >> 33; // shifts 1 bit to the right

An int is 32 bits, so the largest sensible shift would be 31 positions. 31 happens to be the largest number representable by 5 bits. (working with bytes to keep size reasonable) int i = -2; an int is created and initialized to -2, in binary: -2 = 1111 1110 i >>>= -1; right-shift, zero fill, by -1. we know -1 = 1111 1111, but, as we're working with ints, we're only interested in the 'low-order 5 bits'. behind the scenes, a mask of 31 is 'anded' to our shift amount: 1111 1111 &0001 1111 =0001 1111 == 31, so our -1 right shift is really a 31 right shift. -2 >>>= 31 == 0000 0001 == 1 another example int i = 123 << 32; 123 == 0111 1011 32 == 0010 0000 0010 0000 &0001 1111 =0000 0000 123 << 0 == 0111 1011 == 123 Hopefully, that was clear enough. For longs, its the same idea but with 64 bits in the long and 6 bits in the shift. hth