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POMore efficient or more modern? Reading in & Sorting A Text File With Java
 

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    <p>I've been trying to upgrade my Java skills to use more of Java 5 &amp; Java 6. I've been playing around with some programming exercises. I was asked to read in a paragraph from a text file and output a sorted (descending) list of words and output the count of each word.</p> <p>My code is below.</p> <p>My questions are:</p> <ol> <li><p>Is my file input routine the most respectful of JVM resources?</p></li> <li><p>Is it possible to cut steps out in regards to reading the file contents and getting the content into a collection that can make a sorted list of words?</p></li> <li><p>Am I using the Collection classes and interface the most efficient way I can?</p></li> </ol> <p>Thanks much for any opinions. I'm just trying to have some fun and improve my programming skills.</p> <pre><code>import java.io.*; import java.util.*; public class Sort { public static void main(String[] args) { String sUnsorted = null; String[] saSplit = null; int iCurrentWordCount = 1; String currentword = null; String pastword = ""; // Read the text file into a string sUnsorted = readIn("input1.txt"); // Parse the String by white space into String array of single words saSplit = sUnsorted.split("\\s+"); // Sort the String array in descending order java.util.Arrays.sort(saSplit, Collections.reverseOrder()); // Count the occurences of each word in the String array for (int i = 0; i &lt; saSplit.length; i++ ) { currentword = saSplit[i]; // If this word was seen before, increase the count &amp; print the // word to stdout if ( currentword.equals(pastword) ) { iCurrentWordCount ++; System.out.println(currentword); } // Output the count of the LAST word to stdout, // Reset our counter else if (!currentword.equals(pastword)) { if ( !pastword.equals("") ) { System.out.println("Word Count for " + pastword + ": " + iCurrentWordCount); } System.out.println(currentword ); iCurrentWordCount = 1; } pastword = currentword; }// end for loop // Print out the count for the last word processed System.out.println("Word Count for " + currentword + ": " + iCurrentWordCount); }// end funciton main() // Read The Input File Into A String public static String readIn(String infile) { String result = " "; try { FileInputStream file = new FileInputStream (infile); DataInputStream in = new DataInputStream (file); byte[] b = new byte[ in.available() ]; in.readFully (b); in.close (); result = new String (b, 0, b.length, "US-ASCII"); } catch ( Exception e ) { e.printStackTrace(); } return result; }// end funciton readIn() }// end class Sort() ///////////////////////////////////////////////// // Updated Copy 1, Based On The Useful Comments ////////////////////////////////////////////////// import java.io.*; import java.util.*; public class Sort2 { public static void main(String[] args) throws Exception { // Scanner will tokenize on white space, like we need Scanner scanner = new Scanner(new FileInputStream("input1.txt")); ArrayList &lt;String&gt; wordlist = new ArrayList&lt;String&gt;(); String currentword = null; String pastword = null; int iCurrentWordCount = 1; while (scanner.hasNext()) wordlist.add(scanner.next() ); // Sort in descending natural order Collections.sort(wordlist); Collections.reverse(wordlist); for ( String temp : wordlist ) { currentword = temp; // If this word was seen before, increase the count &amp; print the // word to stdout if ( currentword.equals(pastword) ) { iCurrentWordCount ++; System.out.println(currentword); } // Output the count of the LAST word to stdout, // Reset our counter else //if (!currentword.equals(pastword)) { if ( pastword != null ) System.out.println("Count for " + pastword + ": " + CurrentWordCount); System.out.println(currentword ); iCurrentWordCount = 1; } pastword = currentword; }// end for loop System.out.println("Count for " + currentword + ": " + iCurrentWordCount); }// end funciton main() }// end class Sort2 </code></pre>
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    1. COThe first thing that stands out is your C++ background. You may get more out of the exercises if you attempt to make your solutions object-oriented, even if the questions do not specifically ask for it. Making it more object-oriented will get you thinking about how to group functionality into logical classes and hide implementation details behind more-convenient method calls. That said, time to read more of your code and address your question more directly...
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    2. COyour naming conventions are atrocious for modern Java. Hungarian notation that isn't even consistent isn't idiomatic to Java of any version! Directly using `Array` is frowned upon as well, there are `List` and `Set` classes that are more idiomatic as well.
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    3. COJarrod. I understand the comment about Hungarian notation. Why are the List or Set classes better than using an Array in this situation?
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