and btw, as I said earlier, I would prefer to use the largest possible
value, which is likely to be 64K and not maxint. If only one could
point us to the specs for this exact value...

On Tue, Feb 15, 2011 at 2:27 PM, Pierre Joye <[email protected]> wrote:
> why don't you fix it the way you want instead of arguing endlessly? I
> gave all info to explain the reasoning behind that code, if it is
> wrong, and that's totally possible, then fix it and let move on :)
>
> Cheers,
>
> On Tue, Feb 15, 2011 at 2:21 PM, Dmitry Stogov <[email protected]> wrote:
>> On 02/15/2011 03:31 PM, Pierre Joye wrote:
>>>
>>> On Tue, Feb 15, 2011 at 12:54 PM, Dmitry Stogov<[email protected]>  wrote:
>>>
>>>> I don't know all the logic but it looks very strange
>>>>
>>>> 2147483648 == 0x80000000 != 0x7fffffff == 2147483647
>>>>
>>>> so byte_count_signed may be 0x80000000 == -1
>>>
>>> negative value are checked before. Btw, byte_count_signed  is a 64bit
>>> signed integer, for this reason :)
>>
>> Lets byte_count_signed = 0x0000000080000000
>>
>> it's not less than zero and not greater than 2147483648.
>> So the following condition is false.
>>
>> if (byte_count_signed < 0 || (byte_count_signed > 2147483648)) {
>>
>> Then you convert it to size_t.
>>
>> byte_count = (size_t)byte_count_signed;
>>
>> In case size_t is defined as signed int32 you will get byte_count = -1.
>
>
> --
> Pierre
>
> @pierrejoye | http://blog.thepimp.net | http://www.libgd.org
>



-- 
Pierre

@pierrejoye | http://blog.thepimp.net | http://www.libgd.org

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