Re: int32 multiplication, with truncation on overflow
- From: Mike Pall <mike-1405@xxxxxxxxxx>
- To: luajit@xxxxxxxxxxxxx
- Date: Thu, 1 May 2014 11:51:35 +0200
Duncan Cross wrote:
> local bit = require 'bit'
>
> result = tonumber(bit.arshift(bit.lshift((0LL + a) * (0LL + b), 32), 32))
You can drop one widening. And bit.tobit() is the function you
want to use for truncating 64 bit to 32 bit.
The JIT-compiler knows how to turn that all of that into a
truncating 32 bit multiply:
$ luajit -jdump -e "local x = 1; for i=1,201,2 do x = bit.tobit(x*(i+0LL)) end;
print(x)"
...
0bcaffe3 imul ebp, r15d // 32 bit * 32 bit => 32 bit multiply
...
2114520369
--Mike
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