Read and convert binary files to ASCII text

This is my code which receives a file to read and converts binary values to ASCII text.

The code works fine, so I am looking for a review on what I have written, and suggestions on how I could do it better, perhaps without the use of a StringBuffer.

import java.io.File;
import java.util.Scanner;

/**
 *
 * @author Tumpi
 */
public class DN6 {

public static void main(String[] args)
        throws Exception {

    File file = new File("sporocilo.txt");

    Scanner sc = new Scanner(file);
    String lastString = "";

    while (sc.hasNextLine()) {
        String line = sc.nextLine();
        lastString = lastString + line;
    }
        StringBuffer result = new StringBuffer();
        for (int i = 0; i < lastString.length(); i += 8) {
            result.append((char) Integer.parseInt(lastString.substring(i, i + 8), 2));
        }
        System.out.println(result);
    } 

}

Aside from turning StringBuffer into StringBuilder, there are a few more things I'd like to note about your code.

Compound assignment operator

This line:

lastString = lastString + line;

Can be rewritten as this one:

lastString += line;

It's short and it complies with what many people would expect: a concatenation. By explicitly mentioning the original variable again, many people might think it is a different variable before looking at it more closely.

Additionally, you might also want to add a StringBuilder to the reading of the file. Even if the inputfile isn't that big, it would still be consistent.

Magic values

Consider this code:

for (int i = 0; i < lastString.length(); i += 8) {
     result.append((char) Integer.parseInt(lastString.substring(i, i + 8), 2));
}

Can someone other than the authro tell with certainty what the value 8 is? An educated guess would lead to the bit representation of 2 bytes, but we shouldn't have to guess that. Get rid of this so called "magic value" by adding a variable which clears it up:

int amountOfBits = 8;
for (int i = 0; i < lastString.length(); i += amountOfBits ) {
     result.append((char) Integer.parseInt(lastString.substring(i, i + amountOfBits ), 2));
}

Exception handling

You're completely bypassing exception handling. I'm assuming this is just so this contrived sample wouldn't get cluttered with code, but just in case it isn't:

throws Exception

is always a bad idea in an actual program. You should catch the exact exception and handle it appropriately.

StackExchange Code Review Q#46080, answer score: 12