Re: Bignums and .sleb128

[Paul Schlie <schlie@comcast.net>]

(please accept my apologies, and disregard my ramblings)

> From: Paul Schlie <schlie@comcast.net>
> So to be clearer:
> 
>  -1 == -0x1 == -0xF...
> 
> So: assuming signed char:8, short:16, int:32, etc.
> 
>  -1 == (signed char)+0xFF == (short)+0xFFFF == (int)+0xFFFFFFFF
>  -1 == (signed char)-0xFF == (short)-0xFFFF == (int)-0xFFFFFFFF
> 
> and correspondingly:
> 
>    1 == (unsigned char)+0x1        == (unsigned short)+0x1, etc.
>  255 == (unsigned char)-0x1, 65536 == (unsigned short)-0x1, etc.
> 
> As:
> 
>  +0xF == [0...]1111 == 15 (as non-explicitly negative constant is unsigned)
>  -0xF == [1...]1111 == -1 (as  an explicitly negative constant is signed)
> 
> 
>> From: Paul Schlie <schlie@comcast.net>
>>> Daniel Jacobowitz <drow@false.org> writes:
>>>>> You said later that:
>>>>> 
>>>>>> If we're going to use these semantics, at least the '-' case in
>>>>>> operand() needs to be fixed.
>>>>> 
>>>>> but I wasn't sure what you meant by "these semantics".  Do you mean
>>>>> treating bignums as signed, or treating them as unsigned?  By my reading,
>>>>> operand()'s current handling of '-' already assumes they are signed,
>>>>> just like the sleb128 code does (and did ;).
>>>> It doesn't work, because sometimes bignums are signed and sometimes
>>>> they aren't.  Consider -0xffffffffffff; the current code will return 1.
>>>> If you want to treat the input as unsigned, then you need to add a new
>>>> word with the sign bit.  Note that with one less leading 'f', it
>>>> suddenly works.
>> 
>> Strongly suspect that the proper idiom to is to treat all non-explicitly
>> negative constants as being unsigned values; where the point of confusion
>> is that with the exception of decimal numbers; binary, octal, and hex
>> digits directly correspond to N-bit patters which were likely specified
>> as such with the implicit intent they be preserved, the only remaining
>> ambiguity is whether the most-significant specified set-bit is intended
>> to be sign-extended if the value is stored with greater precision than
>> than the otherwise required as determined by the most-significant non-0
>> bit position i.e.:
>> 
>>   -0x1 == [1...]1, where [1...] represents the variable precision
>>   sign-extension of the most significant bit explicitly specified, which
>>   would otherwise only require a signed-bit-field:1, but would need to
>>   be sign extended to fill the remaining most-significant bits if stored
>>   with greater precision as may be required.
>> 
>> Thereby all constant values may be treated uniformly:
>> 
>>   +1 == +0b01 == +0x1 ...
>> 
>>   -1 == -0b01 == -0x1 ...
>> 
>>   +2 == +0b10 == +0x2 ...
>> 
>>   -2 == -0b10 == -0x2 ...
>> 
>> 
>> Which seems quite sensible.
>> 
>>