Re: binary conversion

>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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