Hi, Disregard my previous message, I got your idea =) When we are shifting right then left, then all bits except 7th will be 0 and andmask may be any, while it contains 0x80. Yes everything is ok here, thanks
--- With best regards, Konstantin On Tue, Jul 23, 2013 at 5:05 PM, Konstantin Vladimirov <konstantin.vladimi...@gmail.com> wrote: > Hi, > > But this is awfully wrong. In the general case (value >> 2) & 0xff != > (value >> 2) & 0x80 > > Take value to be 0x3ff for example. Then 0xff != 0x80 itself. This > leads to wrong result. > > --- > With best regards, Konstantin > > On Tue, Jul 23, 2013 at 4:57 PM, David Given <d...@cowlark.com> wrote: >> Konstantin Vladimirov wrote: >> [...] >>> x = (y & ~(1 << 7)) | (((value >> 9) & 1) << 7); >> [...] >>> x = y & 4294967167 | (value >> 9) << 7 & 255; <--------- WAT? >> >> >> ((value >> 9) & 1) << 7 >> == ((value >> 9) << 7) & (1 << 7) >> == ((value >> 9) << 7) & 0x80 >> == ((value >> 9) << 7) & 0xff >> >> ...I think. >> >> That last step is probably being done because anding with 0xff is really >> cheap on x86 (you just pick the appropriate subreg --- al instead of >> eax, for example). >> >> -- >> ┌─── ｄｇ＠ｃｏｗｌａｒｋ．ｃｏｍ ───── http://www.cowlark.com ───── >> │ "USER'S MANUAL VERSION 1.0: The information presented in this >> │ publication has been carefully for reliability." --- anonymous >> │ computer hardware manual >>
