Java - overwrite portion of binary file

Question

I am trying to overwrite last 26 bytes of file. Basically i need to put a few integer and byte variables there. I'm trying to use a DataOutputStream together with FileOutputStream, but these things do not have a seek() method or something like it. So how could i do a writeInt() starting at (file size - 26)? I see there is a write method that accepts an offset but i'm not sure if it is what i want and if so, how to convert int, long and byte variables into byte[] to pass into that method.

Thank you for your advice

WhiteFang34 · Accepted Answer · 2011-03-15 22:37:06Z

7

Using RandomAccessFile you can do something along these lines:

File myFile = new File (filename);
//Create the accessor with read-write access.
RandomAccessFile accessor = new RandomAccessFile (myFile, "rws");
int lastNumBytes = 26;
long startingPosition = accessor.length() - lastNumBytes;

accessor.seek(startingPosition);
accessor.writeInt(x);
accessor.writeShort(y);
accessor.writeByte(z);
accessor.close();

I hope it helps! Let me know if it is good enough or not.

edited Mar 15, 2011 at 22:37

WhiteFang34

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answered Mar 15, 2011 at 21:43

Kaushik Shankar

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2 Comments

Roman Zenka Over a year ago

He might use the accessor.writeInt(int) kind of methods directly without packaging everyting into a byte array.

Marius Over a year ago

Thank you for the nice example, RandomAccessFile is what i was looking for.

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