Author: Uri Blass
Date: 09:59:57 02/08/04
Go up one level in this thread
On February 08, 2004 at 12:27:23, Sune Fischer wrote:
>On February 08, 2004 at 10:29:19, Uri Blass wrote:
>
>>On February 08, 2004 at 09:59:37, Sune Fischer wrote:
>>
>>>>
>>>>Under MSVC, test the following:
>>>>
>>>> unsigned __int64 a = (1 << 50);
>>>> printf("%I64u\n", a);
>>>>
>>>>What will the result be?... 0 ! No errors, no warnings.
>>>
>>>I think the compiler sees the code as:
>>>
>>>int one = 1;
>>>__int64 a = one<<50;
>>>
>>>so the behavior is expected, no warning should be given IMO, there are plenty of
>>>cases where you deliberately want to over shift something to produce a zero.
>>
>>I disagree.
>>You could claim that the compiler should never give a warning by this logic.
>>
>>The behaviour is clearly not expected for the programmer who writes that
>>code(otherwise the programmer could write simply __int64 a=0;
>
>Perhaps a warning wouldn't hurt, although I'm not sure the compiler in general
>will analyze mathematical expressions to see if they produce something
>illogical.
>
>For instance if you have x=a*c*0*d/2; do you expect the compiler to give a
>warning here or should it just compute x=0?
I exopect the compiler to give a warning.
>
>Suppose you have a table of constants, some of which enter into the complex
>expression and are zero. In principle the compiler should be able to unroll the
>loops and give the same warnings.
>
>>>>After debugging for several hours I finally found the problem and fixed it
>>>>using:
>>>>
>>>> unsigned __int64 a = (1ui64 << 50);
>>>>
>>>>Yes, 1 should be written as 1ui64, or another solution:
>>>>
>>>> unsigned __int64 a = (((unsigned __int64) 1) << 50);
>>>>
>>>>Casting 1 to 64 bit.
>>>
>>>You must do the cast before the operation, ie. if you do something like:
>>>
>>>int x=...,y=...;
>>>__int64 f = x*y;
>>>
>>>it will not produce a __int64 result, similarly for a shift operation.
>>
>>Yes and I think that it is a mistake in defining the standard C behaviour.
>
>I don't think it is a mistake.
>
>If you take two floats and multiply them the result is a float and not a double.
>If you wanted a double result you should have used double oprands (or casted to
>double).
>Multiply a float with a float you get a float, divide an integer with an integer
>you get an integer, an appealing consitancy which is easy to remember.
I think that it is more easy to remember that you always get the mathematical
result of the operation in case that it is possible(of course if you divide 2
integers and put the result in an integer varaible then it is not possible to
get the mathematical result).
If you like to get something different then you should mention it.
Uri
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