4 ms·
Does anyone know why the bitshift method works? Is it that low and high are both floating point, so you're not constrained by int precision and so you don't ge
by dunhuang_nomad 4y ago
Does anyone know why the bitshift method works?
Is it that low and high are both floating point, so you're not constrained by int precision and so you don't get an overflow error. The article makes it sound like sign switching is the issue, but this is just a general overflow problem, right?
- dataflow 4y agoThe ">>>" operator works, the ">>" operator doesn't. The reason the former works is that it basically performs unsigned division by a power of 2; the latter does it signed. There's no floating-point.
- erikpukinskis 4y agoWhat do the five >s and the , mean in this comment?
- dataflow 4y ago>>> is bitwise right shift (fills in with zeros), >> is arithmetic right shift (fills in with the sign bit).
- a1369209993 4y ago> >>> is bitwise right shift Well, they're both bitwise right shifts, the ">>>" is specifically a logical or unsigned right shift.
- dataflow 4y agoWhoops yes I meant logical.
- odo1242 4y agoNo, it's because the reason that integers overflow is that negative numbers are technically stored as larger than positive numbers in the Two's complement representation most computers use to store integers. Neither low and high are floats. Example with 8-bit integers (from wikipedia): Bits, Unsigned value, Signed value 0000 0000, 0, 0 0000 0001, 1, 1 0000 0010, 2, 2 0111 1110, 126, 126 0111 1111, 127, 127 1000 0000, 128, −128 When the logical bit shift is conducted on -128, -128 is treated as an unsigned integer. Its sign bit gets shifted such that the integer becomes 0100 0000, aka 64.
- dunhuang_nomad 4y agoOh I see, this is very helpful. Thank you.