I'm reading a binary file like this:
InputStream in = new FileInputStream( file );
byte[] buffer = new byte[1024];
while( ( in.read(buffer ) > -1 ) {
int a = // ???
}
What I want to do it to read up to 4 bytes and create a int value from those but, I don't know how to do it.
I kind of feel like I have to grab 4 bytes at a time, and perform one "byte" operation ( like >> << >> & FF and stuff like that ) to create the new int
What's the idiom for this?
EDIT
Ooops this turn out to be a bit more complex ( to explain )
What I'm trying to do is, read a file ( may be ascii, binary, it doesn't matter ) and extract the integers it may have.
For instance suppose the binary content ( in base 2 ) :
00000000 00000000 00000000 00000001
00000000 00000000 00000000 00000010
The integer representation should be 1 , 2 right? :- / 1 for the first 32 bits, and 2 for the remaining 32 bits.
11111111 11111111 11111111 11111111
Would be -1
and
01111111 11111111 11111111 11111111
Would be Integer.MAX_VALUE ( 2147483647 )
Answer
ByteBuffer has this capability, and is able to work with both little and big endian integers.
Consider this example:
// read the file into a byte array
File file = new File("file.bin");
FileInputStream fis = new FileInputStream(file);
byte [] arr = new byte[(int)file.length()];
fis.read(arr);
// create a byte buffer and wrap the array
ByteBuffer bb = ByteBuffer.wrap(arr);
// if the file uses little endian as apposed to network
// (big endian, Java's native) format,
// then set the byte order of the ByteBuffer
if(use_little_endian)
bb.order(ByteOrder.LITTLE_ENDIAN);
// read your integers using ByteBuffer's getInt().
// four bytes converted into an integer!
System.out.println(bb.getInt());
Hope this helps.