Getting the names of the first n files without listing all of them in linux

Question

I know it is possible to list the first n files in a directory by simply typing: ls | head -n where n is the number of files to be listed. However the problem is the directory I'm trying to run this at is full with thousands of files causing this action to be very slow (takes a couple of minutes) while all I wanted was just the names of the first files.

Is there any way of achieving the desired behavior? and how?

Answer 1

If you don't mind about the ordering of the files..

You need to call the underlying libc opendir/readdir functions as this perl example shows

#!/usr/bin/perl

opendir(my $dh, ".") || die "can't opendir . : $!";
for (1..10) {
    $filename=readdir($dh);
    print "$filename\n";
}

or as a oneliner, which you could call from a shell script

perl -le 'opendir($d, "."); print scalar readdir($d) for (1..10);'

ls -l |head gives files in alphabetic order though which this cannot do. All the files have to be read to get alphabetic ordering