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Re: Bignums and .sleb128
- From: Paul Schlie <schlie at comcast dot net>
- To: Daniel Jacobowitz <drow at false dot org>,Richard Sandiford <rsandifo at redhat dot com>
- Cc: <binutils at sources dot redhat dot com>
- Date: Tue, 01 Feb 2005 00:03:54 -0500
- Subject: Re: Bignums and .sleb128
(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.
>>
>>