Java. quickest way to code to load a file

Question

How in the shortest amount of lines can you load a file of N lines into an ArrayList of strings.

This is what I have, anyone have any suggestions for how to cut down line count and objects?

import java.io.BufferedReader;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.util.ArrayList;

public class FileLoad {

    public static void main(String[] args) throws IOException, FileNotFoundException {
        List<String> hs = new ArrayList<String>();
        BufferedReader br = new BufferedReader(new FileReader(args[0]));
        String line;
        while ((line = br.readLine()) != null) {
            hs.add(line);
        }
    }
}

IMHO this question belongs to http://codereview.stackexchange.com/ — BackSlash, Aug 12 '13 at 13:17
I'll suggest you to use the interface `List` instead of the implementation `ArrayList`. More information here http://stackoverflow.com/questions/147468/why-should-the-interface-for-a-java-class-be-prefered — Marc-Andre, Aug 12 '13 at 13:28

score 5 · Accepted Answer · answered Aug 12 '13 at 13:17

5

In Java 7+, one line:

List<String> lines = Files.readAllLines(Paths.get(args[0]), Charset.forName("UTF-8"));

answered Aug 12 '13 at 13:17

assylias

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score 1 · Answer 2 · answered Aug 12 '13 at 13:21

1

You can use FileUtils.readLines(File f) method from Apache Commons IO Jar

List<String> lines  = FileUtils.readLines(new File("readme.txt"));

answered Aug 12 '13 at 13:21

sanbhat

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Java. quickest way to code to load a file

2 Answers2