OT Ceckarsky kviz II

[Jan Waclawek]

Hmmmm musime najst nejake kompromisne miesto na pitie toho piva.

Mali by sme spriemerovat nejake suradnice alebo tak nejak, len aby nam nepretiekli premenne ;-)


wek


----- Original Message ---------------

Subject: Re: OT Ceckarsky kviz II
   From: =?ISO-8859-2?Q?Miroslav_=A9inko?= <sinkomiro@gmail.com>
   Date: Thu, 5 Mar 2009 19:31:04 +0100
     To: HW-news <hw-list@list.hw.cz>

>Ono sa nielen ze strati znamienko, ono sa cislo sa bude chapat presne
>podla jeho reprezentacie, kym bolo este znamienkove. T.j. -50 sa pri
>32-bit integeri zmeni  4294967246.
>
>Teraz ma od nas po pive aj Lukas :-)
>
>miro
>
>2009/3/5, Jan Waclawek <konfera@efton.sk>:
>> Kua, to akoze sa to znamienko strati?
>>
>> Mnojo, vidim, ze tych piv este bude... ;-)
>>
>> wek
>>
>> ----- Original Message ---------------
>>
>> >Uzasne! Toto by ma pri pisani programu nenapadlo...Vysvetlenie sa ale
>> >skryva v inom pravidle:
>> >
>> >Otherwise, if the operand that has unsigned integer type has rank greater or
>> >equal to the rank of the type of the other operand, then the operand with
>> >signed integer type is converted to the type of the operand with unsigned
>> >integer type.
>> >
>> >miro
>> >
>> >2009/3/5, Luká¹ Grepl <L.Grepl@sh.cvut.cz>:
>> >> > If an int can represent all values of the original type, the value is
>> >> > converted to an int;
>> >> > otherwise, it is converted to an unsigned int. These are called the integer
>> >> > promotions.48) All other types are unchanged by the integer promotions.
>> >>
>> >> Tohle to má urèité vtipné dùsledky. Zrovna pøed pár dny jsem se s tím
>> >> pìknì vypekl a to jsem si (naivka) myslel, ¾e u¾ mi integral promotion
>> >> nedìlá problémy.
>> >>
>> >> Tak¾e kdy¾ u¾ jsme u tìch kvízù, jakých hodnot mù¾e nabývat c?
>> >>
>> >>        int a = -50;
>> >>        unsigned int b = 50;
>> >>
>> >>        int c = a * b / 10;
>> >>
>> >> Luká¹ Grepl
>> >>
>> >> P.S. Integral promotion: It doesn't have to be logical. It's the law.
>>
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