Ah thanks, for all the suggestions! @Eric On my machine the code_native is
the same as Steven's (boils down to single bitshift):
.text
Filename: In[15]
Source line: 1
pushq %rbp
movq %rsp, %rbp
shrq $63, %rdi
Source line: 1
movb %dil, %al
popq %rbp
ret
On Sunday, August 14, 2016 at 12:00:53 AM UTC+2, Stefan Karpinski wrote:
>
> Another way to test it would be
>
> firstbit(n::Integer) = n != (n << 1) >> 1
>
>
> This does not count negative integers. Or if you want to count negative
> integers, then this does the trick:
>
> firstbit(n::Integer) = n != (n << 1) >>> 1
>
>
> For all of the above, calling this on BigInts doesn't make much sense
> since they don't have a high bit, but if you did want to be able to do
> that, you'd probably want to specialize it like this:
>
> firstbit(n::BigInt) = false
>
>
>
> On Sat, Aug 13, 2016 at 3:30 PM, Erik Schnetter <[email protected]
> <javascript:>> wrote:
>
>> You can also write
>> ```Julia
>> firstbit(n::Integer) = signed(n) < 0
>> ```
>>
>> I hope that LLVM generates the same machine code for this as for Steven's
>> bit masks above. Using bit masks is more general since it allows you to
>> test every bit, not just the "first" (sign) bit.
>>
>> -erik
>>
>>
>> On Sat, Aug 13, 2016 at 11:10 AM, Steven G. Johnson <[email protected]
>> <javascript:>> wrote:
>>
>>>
>>>
>>> On Saturday, August 13, 2016 at 9:22:14 AM UTC-4, jw3126 wrote:
>>>>
>>>> Sorry my question was confusing. What I want is a function that behaves
>>>> as follows on bittypes:
>>>>
>>>> *0*0101101 -> 0
>>>> *0*1110101 -> 0
>>>> *0*1011011 -> 0
>>>> *1*0011101 -> 1
>>>> *1*1010001 -> 1
>>>> *1*0110111 -> 1
>>>>
>>>
>>> Normally to compute this function you would just check whether the
>>> bitwise-and with 10000000 is nonzero. e.g.
>>>
>>> firstbit(n::Union{UInt8,Int8}) = (n & 0x80) != 0
>>> firstbit(n::Union{UInt16,Int16}) = (n & 0x8000) != 0
>>> firstbit(n::Union{UInt32,Int32}) = (n & 0x80000000) != 0
>>> firstbit(n::Union{UInt64,Int64}) = (n & 0x8000000000000000) != 0
>>> firstbit(n::Union{UInt128,Int128}) = (n &
>>> 0x80000000000000000000000000000000) != 0
>>> firstbit(n::Integer) = leading_ones(n) > 0
>>>
>>
>>
>>
>> --
>> Erik Schnetter <[email protected] <javascript:>>
>> http://www.perimeterinstitute.ca/personal/eschnetter/
>>
>
>
