macOS Can you guys explain to me a certain part of this code.

[slavegod]

use strict;
use warnings;

my @list = (4,5,6,7,1,2,3);
my @x;
my @y;
my @z;

my $bin = 0;

foreach my $e ( @list ) {
if ( $bin == 2){
unshift @z, $e;
$bin = 0;
} elsif ( $bin ) {
push @y, $e, $e+1;
$bin = 2;
} else {
push @x, $e;
$bin = 1;
}
}

print "x is: @x\n";
print "x is: @y\n";
print "z is: @z\n";


this is perl. I just don't get the part from my $bin ; to the end of the foreach loop... can someone please explain. Thanks.

 

[angelwatt]

What parts don't you get, the loop, the conditions, the shifting and unshifting? Try running the program, it may help you understand the code. It's pretty simple code.

 

[lee1210]

I don't know what we could tell you that you wouldn't find out running it yourself.

For each three numbers in @list, the first is added to the end of @x, the second is added to the end of @y, then that value plus one is added to the end of @y, the third is added to the beginning of @z.

-Lee

 

[slavegod]

i just don't get what the $bin is for?? and i ran the code but i just don't really get it.


x is: 4 7 3
y is: 5 6 1 2
z is: 2 6


can someone just explain to me why x is as is.. because this is a practice exam and my prof is kinda bad at explaining...

 

[autorelease]

$bin indicates which array the number from @list gets put into. 0 indicates @x, 1 indicates @y, and 2 indicates @z.

If $bin is 0, $e gets added to the end of @x.
If $bin is 1, $e and $e+1 get added to the end of @y.
If $bin is 2, $e gets added to the beginning of @z.

The order of the conditions is kind of weird. It might make more sense rearranged like this:

foreach my $e ( @list ) {
  if ($bin == 0) {
    push @x, $e;
    $bin = 1;
  } elsif ($bin == 1) {
    push @y, $e, $e+1;
    $bin = 2;
  } elsif ($bin == 2) {
    unshift @z, $e;
    $bin = 0;
  }
}

# (Since $bin will always be 0, 1, or 2, this functions the same as the original code despite the rewritten conditions.)

I haven't written Perl in years, looking at this code brought back bad memories...

 

[lee1210]

i just don't get what the $bin is for?? and i ran the code but i just don't really get it.


x is: 4 7 3
y is: 5 6 1 2
z is: 2 6


can someone just explain to me why x is as is.. because this is a practice exam and my prof is kinda bad at explaining...

$bin is for determining if you're at the 1st, 2nd, or 3rd position of the current set of three.. or put differently, whether the position modulo 3 of @list is 0,1, or 2.

Is is as it is because x has the 1st element of each set of three numbers in @list added to the end.

I think it would be best to step through the code bit by bit... this time around I did it for you, but you can do this yourself or you can add print statements to see what's going on each step of the way:

my $bin = 0;
foreach $e ( @list ) { //First run, $e is 4
…
  } else { // because bin is 0
    push @x, 4; // @x is now (4)
    $bin=1
  }
}

foreach $e ( @list ) { //Second run, $e is 5, $bin is 1
  } elsif ( $bin) { // $bin is not 2, but it is non-zero
    push @y, 5, 6; // @y is (5,6)
    $bin = 2;
  }
}

foreach $e ( @list ) { //Third run, $e is 6, $bin is 2
  if( $bin == 2) { //Since $bin is 2
    unshift @z, 6; //@z is now (6)
    $bin = 0;
  }
}

foreach $e ( @list ) { //Fourth run, $e is 7, $bin is 0
…
  } else { // because bin is 0
    push @x, 7; // @x is now (4,7)
    $bin=1
  }
}

foreach $e ( @list ) { //Fifth run, $e is 1, $bin is 1
  } elsif ( $bin) { // $bin is not 2, but it is non-zero
    push @y, 1, 2; // @y is (5,6,1,2)
    $bin = 2;
  }
}

foreach $e ( @list ) { //Sixth run, $e is 2, $bin is 2
  if( $bin == 2) { //Since $bin is 2
    unshift @z, 2; //@z is now (2,6)
    $bin = 0;
  }
}

foreach $e ( @list ) { //Seventh run, $e is 3, $bin is 0
…
  } else { // because bin is 0
    push @x, 3; // @x is now (4,7,3)
    $bin=1
  }
}

-Lee

 

[slavegod]

Man finally figured it out... Thanks guys