Re: Bitwise shift operator on 8-byte integers

[Ankit Jain <ankitjain1580 at yahoo dot com>]

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@iinet.net.au> 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@siemens.com 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
> 
>  

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