thanks a lot so its basically a problem of type casting well i have 1 more ques i have a signed 32bit integer var var=1; var<<=31; var>>=31; var displays -1 here and after this if i do var>>=1; it still diplays -1. i could not understand why on right shift the sign bit is not changed to 0 whats happening here? thanks ankit --- Sisyphus <kalinabears@xxxxxxxxxxxx> wrote: > Ankit Jain wrote: > > hi Eljay, > > > > I understand the solution given by u and its > workign > > is also fine . but could not understand why this > is > > not working > > > > thanks > > > > ankit > > --- Krzysztof.Wisniowski@xxxxxxxxxxx wrote: > > > >>Hallo *, > >>Here's the problem: > >> > >>unsigned int ui = 4294967295; //2^16-1 > > Actually, that's 2^32-1. > > >>unsigned long long uL; //8-byte variable > >> > >>uL = ui << 16; > > You're asking that a 32 bit value (ui) be left > shifted 16 places. This > means that the 16 high bits will be discarded. The > result of that left > shift will then be assigned to uL. > > I believe that another solution would be to write it > as: > uL = (unsigned long long)ui << 16; > > this casts ui to an unsigned long long - so we're > now asking that a 64 > bit value be left shifted 16 places - and no bits > will be lost. > Consequently uL will contain the value you expect. > > Get the picture ? > > Cheers, > Rob > > ________________________________________________________________________ Yahoo! Messenger - Communicate instantly..."Ping" your friends today! Download Messenger Now http://uk.messenger.yahoo.com/download/index.html
