[dunhuang_nomad]

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]

The ">>>" 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]

What do the five >s and the , mean in this comment?

[dataflow]

>>> is bitwise right shift (fills in with zeros), >> is arithmetic right shift (fills in with the sign bit).

[a1369209993]

> >>> is bitwise right shift

Well, they're both bitwise right shifts, the ">>>" is specifically a logical or unsigned right shift.

[dataflow]

Whoops yes I meant logical.

[odo1242]

No, 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]

Oh I see, this is very helpful. Thank you.