OT Ceckarsky kviz II

[Miroslav Šinko]

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.
>