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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