How to get EOL character of any file in java

Question

I am trying to find the EOL character of a file. I have tried the following snippet:

java.io.FileInputStream inputStream = null;  
inputStream=new java.io.FileInputStream (inputFile.trim());
java.util.Properties p = new java.util.Properties(System.getProperties());  
p.load(inputStream);  
System.setProperties(p);  
String lineSeperator=System.getProperties().getProperty("line.separator");  
System.out.println("--"+lineSeperator);

The result i am getting here is always \r\n, Even if i am reading a Linux originated file where delimiter should be \n.

What am i doing wrong ?

Answer 1

System.getProperty("line.separator") returns the default line separator of your operating system. It has nothing to do with what the line separator used by a given file is.

To find it, open the file as a character stream, and iterate over the characters until you find '\r' or '\n'. If you find '\n' first, the you have a unix file. If you find '\r' followed by '\n', you have a windows file. If you find '\r' not followed by '\n', you have an old Mac file.

This of course assumes that the file is a text file in the first place, that a single kind of EOL is used in the file, and that you know the encoding of the file. If you don't know the encoding, you might use the same algorithm and use a byte stream instead of a character stream.

Note that this is irrelevant, most of the time, since Java has APIs that read a file line by line, and handle all combinations of EOLs just fine.