Re: binary conversion

Try this ...

    int num [] = new int[4];

    num[0] = Integer.parseInt("00000000", 2);
    num[1] = Integer.parseInt("00001110", 2);
    num[2] = Integer.parseInt("00110101", 2);
    num[3] = Integer.parseInt("10000001", 2);

    int bigE = ((num[3]) | ((num[2]) << 8) | ((num[1]) << 16) | ((num[0]) << 24));

    System.out.println("bigE = " + bigE);

    // Output: bigE = 931201

Erling

----- Original Message ----- 
From: "K D" <[email protected]>
To: <[email protected]>
Sent: Friday, May 23, 2003 7:38 PM
Subject: Re: binary conversion


> Hi,
> 
> I tried your method, and I got "0" for both way
> I am not sure why,
> but here is my input
> 
> 00000000
> 00001110
> 00110101
> 10000001
> 
> these number = 931201 (year,month,day)
> 
> thanks
> 
> 
> --- Niall Gallagher <[email protected]> wrote:
> > >In the example that you gave it,
> > >you are converting an integer
> > >into big and small endian right?
> > >
> > >I guess what I tried to ask was
> > >I have a 4 byte of data,
> > >and it is not stored in ascci format
> > >but rather in byte data only, so I have
> > >to read in data as byte array like this...
> > >
> > >byte num [] = new byte[4]
> >
> >
> > Hi
> >
> > What you have is a list of 8-bit bytes and what you
> > want to do
> > is to convert these into a single integer. So for
> > example if
> > the integer was 1 you have the equivelant of
> >
> > a [0000 0001]
> > b [0000 0000]
> > c [0000 0000]
> > d [0000 0000]
> >
> > Where a is num[0] and d is num[3]. Well if I
> > remember big
> > endian right what you want is the binary 32-bit
> > integer
> >
> >  [0000 0000 0000 0000 0000 0000 0000 0001]
> >
> > So to do this the following expression should work.
> >
> > int bigE =   ((num[0] & 0x00)
> >        | ((num[1] & 0x00) << 8)
> >        | ((num[2] & 0x00) << 16)
> >        | ((num[3] & 0x00) << 24));
> >
> > This should give you the correct integer value,
> > However
> > I may have my big and little endian upside down, in
> > which
> > case the following sould work
> >
> > int bigE =  ((num[3] & 0x00)
> >        | ((num[2] & 0x00) << 8)
> >        | ((num[1] & 0x00) << 16)
> >        | ((num[0] & 0x00) << 24));
> >
> >
> > I have had to do this many times before, so you
> > should
> > have no problem with these expressions, if you do,
> > let me know.
> >
> > Niall
> >
> >
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