How to find the size of the largest file in a directory?

Question

I need to find out the size of the largest file/folder in a directory. I have done it in the below way.

private static Long getSizeofLargestFile(String theRootFolder)
    {
        Long aLargestFileSize = 0L;
        File aRootDir = new File(theRootFolder);
        for (File aFile : aRootDir.listFiles())
        {
            if (aLargestFileSize < aFile.length())
            {
                aLargestFileSize = aFile.length();
            }
        }
        return aLargestFileSize / (1024 * 1024);
    }

Can there be a better way than this?

Answer 1

Not really.. You have to get the sizes of all files to find the largest, which is what you are doing.

The only thing I would advise against is the division in the return. If the largest file is less than one MB, then your method will return 0!

Answer 2

I would say instead of

return aLargestFileSize / (1024 * 1024);

go with

return aLargestFileSize;

I will write why in some time...

import java.util.*;
import java.lang.*;

class Main
{
    public static void main (String[] args)
    {
        Long aLargestFileSize1 = (1024*1024)+2 + 0L;
        Long aLargestFileSize2 = (1024*1024)+ 0L;

        System.out.println("aLargestFileSize1 final is " + aLargestFileSize1/ (1024 * 1024));
        System.out.println("aLargestFileSize2 final is " + aLargestFileSize2/ (1024 * 1024));

        System.out.println("aLargestFileSize1 new is " + aLargestFileSize1);
        System.out.println("aLargestFileSize2 new is " + aLargestFileSize2);

        }
}

If you see output both are giving 1. But if you would had System.out.println("aLargestFileSize1 final is " + aLargestFileSize1);, then output would have been,

aLargestFileSize1 new is 1048578
aLargestFileSize2 new is 1048576

means first file is larger.

So, just use actual number in return.

Demo