UNIX for Dummies Questions & Answers

Hello everyone,

on the man page of "magic(5)"
There is explanation
"&, to specify that the value from the file must have set all of the bits that are set in the specified value" .

My question is that what is the difference between '&' and equal operator '=' ? I tested it with file command but I do not understand the work way of it. For example, I prepared a magic.mime named 'mymagic' including only one line that is " 0 ubyte &0x00 filetypededected" and I create a file named 'myfile' setting its first byte to \x4e . when i call 'file -m mymagic myfile' it prints filetypededected so how can it be?

I don't have ubyte on my man page, what system are you using? Ignoring that, &0x00 says to scan 0x00 looking only at the 1 bits. Any 1 bits appearing in 0x00 must be matched by the corresponding bit in the byte under test. Since &0x00 has no 1 bits, all bytes will match it. In contrast, =0x00 means that only a byte of zero matches.

Thank you very much friend it works now. "ubyte" is not writing on man pages but it is mentioned in somewhere in magic.mime database files and ulong, ubelong ... , as well. So "^" operator is vice versa of "&" we care about 0 bits, isn't it?


Aaa, I tested "^" operator for 0 bits this time but I have not idea about it.

In magic file, I inserted 0xd3 which is 11010011 in binary form
and in my file which I will test has 0xb1 on his first byte which is 10110001.
I operate ^ operator as "Or" and "exOr" but i could not get any idea.Could you help for this operator?

"&0x1" means low order bit must be 1
"^0x1" means low order bit must be 0

And you're right, I have ubyte in my files but not on my man page.

Sorry, I do not understand functionality of ^ operator by this sentences
" "^0x1" means low order bit must be 0 "

I posted above and will write again clearly.

In my magic.mime there is only one line
"0 byte ^0xd3 file\ detected"

and in my test file named myfile, the first byte is 0xb1.

0xd3 = 11010011
0xb1 = 10110001

when I write 'file -m magic.mime myfile'
it prints "file detected".

According to your explanation there should be 0 bits on testing byte corresponding 1 bits in magic.mime but in example above there is nothing like that.

Xor is a bitwise operator:
0^0=0
0^1=1
1^0=1
1^1=0

If you have multiple 1 bits it is no longer a simple test.

Code:

0xd3 = 11010011
0xb1 = 10110001
       --------
0x62 = 01100010

It's not real obvious why you would do that. But the result is a match. Do you know C language? If so it is like:

Code:

a=0xd3;
b=0xb1;
if (a ^ b)   printf("match\n");

Make them both 0xb1 and the test will fail.

Code:

$
$ cat /tmp/mymagic
0 byte ^0xb1 filedetected
$ od -tx1 /tmp/x
0000000 b1 0a
0000002
$ file -m /tmp/mymagic /tmp/x
/tmp/x: ISO-8859 text
$

I choose a=0x22 and b=0x32.
In C they match but in magic test not matched. When we think like C language it is similar to not equal but it does not work every time. I am implementing filetype detection in erlang so I need functionalities of magic operators.I still have problem with ^ operator.

Edit:
The problem is solved.
At least one 0 bit should be placed in the test file instead of corresponding 1 bits in magic file.

Thank you very much for your helps.