Author: Uri Blass
Date: 08:07:36 09/10/03
Go up one level in this thread
On September 10, 2003 at 11:04:08, Jaime Benito de Valle Ruiz wrote:
>On September 10, 2003 at 09:36:20, Uri Blass wrote:
>
>>On September 10, 2003 at 09:16:04, Tim Foden wrote:
>>
>>>On September 10, 2003 at 08:45:22, Uri Blass wrote:
>>>
>>>>I found that my program printed the numbers 0 and 1 when it did the following
>>>>code without printing mistake or mistake1.
>>>>
>>>>zobpawnkey[hply]=zobpawn;
>>>>if (nodes==663719)
>>>>{
>>>> if (zobpawnkey[hply]!=zobpawn)
>>>> printf("mistake");
>>>> if ((zobpawnkey[hply]&1)!=(zobpawn&1))
>>>> printf("mistake1");
>>>> printf(" %d %d ",(zobpawn&1),(zobpawnkey[hply]&1));
>>> ^^ ^^ these expect int, not __int64.
>>>
>>>If using Microsoft's compiler system, you should use %I64d instead.
>>>
>>>Cheers, Tim.
>>
>>I understand that %d does not give me the right value but the fact that it
>>prints different numbers for identical numbers still seems strange to me.
>>
>>Uri
>
>Because it is printing different numbers from ONLY one number (zobpawn&1), i.e.
>from the lower and upper 32 bits of it. None of those numbers displayed by the
>printf("%d %d"...) has anything to do with (zobpawnkey[hply]&1).
>Probably, (zobpawn) and (zobpawnkey[hply]) in this example are the same, and
>both equal to 0x0000000100000000. That's why you don't get errors in the first
>place, and why you get your 0 and 1 on the screen!
>I hope it's clear what I say.
>Regards,
>
> Jaime
Yes
It is clear
Thanks.
Uri
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