Parsing Binary Files

Parsing binary files is no different than parsing character-based files except that the “characters” are actually bytes not 16-bit unsigned short unicode characters. From a lexer/parser point of view, there is no difference except that the characters are likely not printable. If you want to match a special 2-byte marker 0xCA then 0xFE, the following rule is sufficient.

MARKER : '\u00CA' '\u00FE' ;

The parser of course would refer to that token like any other token.

Here is a sample grammar for use with the code snippets below.

grammar IP;

file : ip+ (MARKER ip)* ;

ip : BYTE BYTE BYTE BYTE ;

MARKER : '\u00CA' '\u00FE' ;
BYTE : '\u0000'..'\u00FF' ;

Notice that BYTE is using a range operator to match anything between 0 and 255. We can't use character classes like [a-z] naturally because we are not parsing character codes. All character specifiers must have 00 as their upper byte. E.g., \uCAFE is not a valid character because that 16-bit value will never be created from the input stream (bytes only remember).

If there are actual characters like $ or ! encoded as bytes in the binary file, you can refer to them via literals like '$' as you normally would. See '.' in the grammar.

Binary streams

There are many targets now so I‘m not sure exactly how they process text files but most targets will pull in text per the machine’s locale. Much of the time this will mean UTF-8 encoding of text converted to 16-bit Unicode. ANTLR's lexers operate on int so we can handle any kind of character you want to send in that fits in int.

Once the lexer gets an input stream, it doesn't care whether the characters come from / represent bytes or actual Unicode characters.

Let's get a binary file called ips and put it in our resources directory:

public class WriteBinaryFile {
	public static final byte[] bytes = {
		(byte)172, 0, 0, 1, (byte)0xCA, (byte)0xFE,
		(byte)10, 10, 10, 1, (byte)0xCA, (byte)0xFE,
		(byte)10, 10, 10, 99
	};

	public static void main(String[] args) throws IOException {
		Files.write(new File("/tmp/ips").toPath(), bytes);
	}
}

Now we need to create a stream of bytes satisfactory to ANTLR, which is as simple as:

CharStream bytesAsChar = CharStreams.fromFileName("/tmp/ips", StandardCharsets.ISO_8859_1);

The ISO-8859-1 encoding is just the 8-bit char encoding for LATIN-1, which effectively tells the stream to treat each byte as a character. That's what we want. Then we have the usual test rig:

//ANTLRFileStream bytesAsChar = new ANTLRFileStream("/tmp/ips", "ISO-8859-1"); DEPRECATED in 4.7
CharStream bytesAsChar = CharStreams.fromFileName("/tmp/ips", StandardCharsets.ISO_8859_1);
IPLexer lexer = new IPLexer(bytesAsChar);
CommonTokenStream tokens = new CommonTokenStream(lexer);
IPParser parser = new IPParser(tokens);
ParseTree tree = parser.file();
IPBaseListener listener = new MyIPListener();
ParseTreeWalker.DEFAULT.walk(listener, tree);

Here is the listener:

class MyIPListener extends IPBaseListener {
	@Override
	public void exitIp(IPParser.IpContext ctx) {
		List<TerminalNode> octets = ctx.BYTE();
		short[] ip = new short[4];
		for (int i = 0; i<octets.size(); i++) {
			String oneCharStringHoldingOctet = octets.get(i).getText();
			ip[i] = (short)oneCharStringHoldingOctet.charAt(0);
		}
		System.out.println(Arrays.toString(ip));
	}
}

We can't just print out the text because we are not reading in text. We need to emit each byte as a decimal value. The output should be the following when you run the test code:

[172, 0, 0, 1]
[10, 10, 10, 1]
[10, 10, 10, 99]

Custom stream

(ANTLRFileStream was deprecated in 4.7)

If you want to play around with the stream, you can. Here's an example that alters how “text” is computed from the byte stream (which changes how tokens print out their text as well):

/** make a stream treating file as full of single unsigned byte characters */
class BinaryANTLRFileStream extends ANTLRFileStream {
	public BinaryANTLRFileStream(String fileName) throws IOException {
		super(fileName, "ISO-8859-1");
	}

	/** Print the decimal value rather than treat as char */
	@Override
	public String getText(Interval interval) {
		StringBuilder buf = new StringBuilder();
		int start = interval.a;
		int stop = interval.b;
		if(stop >= this.n) {
			stop = this.n - 1;
		}

		for (int i = start; i<=stop; i++) {
			int v = data[i];
			buf.append(v);
		}
		return buf.toString();
	}
}

The new test code starts out like this:

ANTLRFileStream bytesAsChar = new BinaryANTLRFileStream("/tmp/ips");
IPLexer lexer = new IPLexer(bytesAsChar);
...

This simplifies our listener then:

class MyIPListenerCustomStream extends IPBaseListener {
	@Override
	public void exitIp(IPParser.IpContext ctx) {
		List<TerminalNode> octets = ctx.BYTE();
		System.out.println(octets);
	}
}

You should get this enhanced output:

[172(0xAC), 0(0x0), 0(0x0), 1(0x1)]
[10(0xA), 10(0xA), 10(0xA), 1(0x1)]
[10(0xA), 10(0xA), 10(0xA), 99(0x63)]

Error handling in binary files

Error handling proceeds exactly like any other parser. For example, let‘s alter the binary file so that it is missing one of the 0’s in the first IP address:

public static final byte[] bytes = {
	(byte)172, 0, 1, (byte)0xCA, (byte)0xFE, // OOOPS
	(byte)10, 10, 10, 1, (byte)0xCA, (byte)0xFE,
	(byte)10, 10, 10, 99
};

Running the original test case gives us:

line 1:4 extraneous input '.' expecting BYTE
line 1:6 mismatched input 'Êþ' expecting '.'
[172, 0, 1, 0]
[10, 10, 10, 1]
[10, 10, 10, 99]

That 'Êþ' is just to the character representation of two bytes 0xCA and 0xFE. Using the enhanced binary stream, we see:

line 1:4 extraneous input '46(0x2E)' expecting BYTE
line 1:6 mismatched input '202(0xCA)254(0xFE)' expecting '.'
[172(0xAC), 0(0x0), 1(0x1)]
[10(0xA), 10(0xA), 10(0xA), 1(0x1)]
[10(0xA), 10(0xA), 10(0xA), 99(0x63)]