Java reading hex file

Question

As part of a larger program, I need to read values from a hex file and print the decimal values. It seems to be working fine; However all hex values ranging from 80 to 9f are giving wrong values. for example 80 hex gives a decimal value of 8364 Please help.

this is my code :

String filename = "pidno5.txt";
FileInputStream ist = new FileInputStream("sb3os2tm1r01897.032");       
BufferedReader istream = new BufferedReader(new InputStreamReader(ist));
int b[]=new int[160];       
for(int i=0;i<160;i++)
    b[i]=istream.read();
for(int i=0;i<160;i++)
    System.out.print((b[i])+" ");

Just for terminology: You are not reading a hex file (which would be a file containing hexadecimal data coded in ASCII?), but a binary file. — Paŭlo Ebermann
– Paŭlo Ebermann, Commented Jun 14, 2011 at 11:58

Thilo · Accepted Answer · 2011-06-14 11:31:28Z

7

If you were trying to read raw bytes this is not what you are doing.

You are using a Reader, which reads characters (in an encoding you did not specify, so it defaults to something, maybe UTF-8).

To read bytes, use an InputStream (and do not wrap it in a Reader).

answered Jun 14, 2011 at 11:31

Thilo

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1 Comment

jarnbjo Over a year ago

The default encoding is probably some Microsoft code page, as some of them have the € sign on code point 0x80, which is then mapped to Unicode 8364.

Ioannis Pappas · Accepted Answer · 2012-08-05 14:40:24Z

0

You may also use a different encoding:

BufferedReader istream = new BufferedReader(new InputStreamReader(ist, "ISO-8859-15"));

answered Aug 5, 2012 at 14:40

Ioannis Pappas

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