This file is copyright © 2006 Mark Jason Dominus. Unauthorized distribution in any medium is absolutely forbidden.

# Inserting Commas

Our next example is a single subroutine that takes a numeral, such as 1234567.89, and returns a version that has commas inserted, as "1,234,567.89". Before we look at the subroutine itself, we should note that this is a Frequently Asked Question; the solution given in the Perl FAQ list is:

```        s/(^[-+]?\d+?(?=(?>(?:\d{3})+)(?!\d))|\G\d{3}(?=\d))/\$1,/g;
```

This is the kind of thing that gives Perl a bad reputation. Let's look at the code we have and see if we can't come up with something a little less ridiculous.

The original function, conversion(), is 30 lines long. The author told me he thought that it seemed bulky. I agree.

The first thing that jumps out at me is that \$pos is an array length variable, so we can easily get rid of it, by replacing this:

```    13	        \$section[\$pos] = substr(\$number, 0, \$remain);
16	        \$pos++;
```

with this:

```        push @section, substr(\$number, 0, \$remain);
```

and this:

```    18	      \$section[\$pos] = substr(\$number, \$next, 3);
20	      \$pos++;
```

with this:

```        push @section, substr(\$number, 0, 3);
```

The main loop now looks like this:

```               while (\$next < \$loop) {
if (\$remain > 0) {
push @section, substr(\$number, 0, \$remain);
\$next = \$remain++;
\$remain = 0;
}
push @section, substr(\$number, \$next, 3);
\$next = (\$next + 3);
}
```

The thing to notice here is the \$remain variable. If the if block is executed, then \$remain is set to zero, and since it is not modified anywhere else, it must remain zero until the loop completes. That means that the if block can execute at most once, and only on the first pass through the loop. So we can hoist it out:

```              if (\$remain > 0) {
push @section, substr(\$number, 0, \$remain);
\$next = \$remain++;
\$remain = 0;
}
while (\$next < \$loop) {
push @section, substr(\$number, \$next, 3);
\$next = (\$next + 3);
}
```

Now instead of one complicated block with nested logic, we have two simple blocks.

## Mumble something

The goal of the while loop is now apparent: it is collecting the three-character substrings of \$number. Armed with this logic, we can make sense of the if block too: it's acquiring the leftover characters from the front of the numeral. In the example of 1234567, the if block acquires the 1 and the while loop acquires 234 and then 567.

```        sub conversion
{
\$number = shift;
\$size = length(\$number);
\$result = (\$size / 3);
@commas = split (/\./, \$result);
\$remain = (\$size - (\$commas[0] * 3));
\$pos = 0;
\$next = 0;
\$loop = (\$size - \$remain);
while (\$next < \$loop)
{
if (\$remain > 0)
{
\$section[\$pos] = substr(\$number, 0, \$remain);
\$next = \$remain++;
\$remain = 0;
\$pos++;
}
\$section[\$pos] = substr(\$number, \$next, 3);
\$next = (\$next + 3);
\$pos++;
}
\$loop = 0;
@con = ();
foreach (@section)
{
\$loop++;
\$cell++;
@tens = split (/:/, \$_);
\$con[\$cell] = \$tens[0];
if (\$loop == \$pos)
{
last;
}
\$cell++;
\$con[\$cell] = ",";
}
return @con
```

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