RogerBW's Blog

Perl Weekly Challenge 98: Insert N 05 February 2021

I’ve been doing the Perl Weekly Challenges. The latest involved reading tiny chunks from a file and searching a list. (Note that this is open until 7 February 2021.)

TASK #1 › Read N-characters

You are given file $FILE.

Create subroutine readN($FILE, $number) returns the first n-characters and moves the pointer to the (n+1)th character.

Task 1 is usually the easier of the two, and indeed this would be fairly trivial, except for the idea that the sub should take the filename as a parameter. Which means that somewhere we need to keep track of how much has already been read, either using an open file handle or with an internal counter; I chose the file handle. But this is a really ugly way of doing things; an object would be much more usual, which you'd initialise with the filename. (Indeed, you might call it "a file handle" since that's how they work.)

In Perl I can define the static variables in the same block as the sub, so that's easy.

{
  my $fh;
  my $fn='';
  sub readN {
    my $fnn=shift;
    my $siz=shift;
    if ($fnn ne $fn) {
      $fn=$fnn;
      if (defined $fh) {
        close $fh;
        undef $fh;
      }
    }
    if (!defined $fh) {
      open $fh,'<',$fn;
    }
    my $buf;
    my $l=read ($fh,$buf,$siz);
    if ($l < $siz) {
      close $fh;
      undef $fh;
      $fn='';
    }
    return $buf;
  }
}

And Raku works similarly, though with state variables (Perl5 has these too, I just didn't use them).

sub readN($fnn,$siz) {
  state $fh=0;
  state $fn='';
  if ($fnn ne $fn) {
    $fn=$fnn;
    if ($fh) {
      $fh.close;
      $fh=0;
    }
  }
  unless ($fh) {
    $fh=$fn.IO.open;
  }
  my $buf=$fh.read($siz).decode;
  if ($buf.chars < $siz) {
    $fh.close;
    $fh=0;
    $fn='';
  }
  return $buf;
}

In Python and Ruby I used a global hash table of filename to handle.

rn=dict()

def readN(fn,n):
    if fn in rn:
        fh=rn[fn]
    else:
        fh=io.FileIO(fn)
        rn[fn]=fh
    buf=fh.read(n)
    if len(buf)==0:
        fh.close
        del rn[fn]
    return buf

and

$rn=Hash.new

def readN(fn,siz)
  if $rn.has_key?(fn) then
    fh=$rn[fn]
  else
    fh=File.new(fn,'r')
    $rn[fn]=fh
  end
  buf=fh.read(siz)
  if fh.eof? then
    fh.close
    $rn.delete(fn)
  end
  return buf
end

and in Rust, after a lot of back-and-forth with unsafe code, I just gave up and made an object out of it the way I would in the real world.

pub struct Reader {
    handle: std::fs::File
}

impl Reader {

    pub fn new(fname: &str) -> Reader {
        let fna=fname.to_string();
        let fh=std::fs::File::open(fna).unwrap();
        Reader { handle: fh }
    }

#[allow(non_snake_case)]

(this to suppress the warning about the capitalised subroutine name)

    pub fn readN(&mut self, siz: usize) -> String {
        let mut buf=vec![0u8; siz];
        let mut taker=self.handle.borrow().take(siz as u64);
        let t=taker.read(&mut buf).unwrap();
        let q=&buf[0..t];
        let s=str::from_utf8(&q).unwrap();
        return s.to_string();
    }
}

TASK #2 › Search Insert Position

You are given a sorted array of distinct integers @N and a target $N.

Write a script to return the index of the given target if found otherwise place the target in the sorted array and return the index.

I skipped the "place the target in the sorted array" bit since the array is immediately thrown away at function exit…

In the real world at this scale I would normally just transmute the array to a set to test for presence (caution, untested code):

my %n=map {$_ => 1} @N;
if (exists $n{$N}) {
  return scalar @N;
}

Then count the smaller entries:

my @t=grep {$_ < $N} @N;
return scalar @t;

which doesn't even need the array to be sorted, but this is clearly a computer science homework problem so I'll try to solve it like one with a binary search.

sub sip {
  my $n=shift;
  my $t=shift;
  if ($n->[-1] < $t) {
    return scalar @{$n};
  }
  my ($l,$h)=(0,$#{$n});
  while ($h-$l > 1) {
    my $m=int(($h+$l)/2);
    if ($n->[$m] == $t) {
      return $m;
    } elsif ($n->[$m] > $t) {
      $h=$m;
    } else {
      $l=$m;
    }
  }
  if ($n->[$l] >= $t) {
    return $l;
  }
  return $h;
}

And the other languages are basically identical.

Full code on github.

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