Shift operator, unexpected result
tsbockman
thomas.bockman at gmail.com
Wed Jun 9 22:03:33 UTC 2021
On Wednesday, 9 June 2021 at 19:13:10 UTC, JG wrote:
> I found the following behaviour, as part of a more complicated
> algorithm, unexpected. The program:
>
> import std;
> void main()
> {
> int n = 64;
> writeln(123uL>>n);
> }
>
> produces:
>
> 123
>
> I would expect 0.
>
> What is the rationale for this behaviour or is it a bug?
Because it is a high-performance systems programming language,
the designers of D decided to make the arithmetic operations of
basic types map directly to the arithmetic operations built in to
the CPU; most operations are a single instruction.
The benefit of this is higher performance and smaller binaries.
The disadvantage is that the behaviour of the built in CPU
operations sometimes differs from ordinary arithmetic in
surprising and frustrating ways.
If you want to trade a some speed for correctness/predictability,
try my `checkedint` Dub package. Either way, take a glance at the
[introduction to the
documentation](https://checkedint.dpldocs.info/checkedint.html),
where I list some of the quirks of CPU integer behaviour.
For bit shifts, specifically, many CPUs ignore all but the bottom
`log2(T.sizeof * 8)` bits of the right-hand operand.
(`core.bitop.bsr` can be used to do very fast integer `log2`
operations, and works in CTFE.) Thus, `a >> b` behaves like `a >>
(b & c)`, where `c` is `(T.sizeof * 8) - 1`.
For unsigned types, the behaviour that you very reasonably expect
requires two additional instruction on x86, which looks like
this: `(b <= c)? (a >> b) : 0`. (This should be branchless thanks
to the `cmov` instruction.)
For signed types, some additional work is required to handle
negative shifts; see my `checkedint` package.
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