Since when is 31 not an integer?

In the description it says that the first variable ($calendar) is an integer, but 
in Example 1 it’s clearly a string. And when you run the code, neither a string or 
integer seems to work.

Just tested on a different server running 5.2.8, it accepts a string or an 
integer! So the code does work, although can’t seem to get it to work on my 5.3.5 
server, but the Example does not follow the Description.

Ah, for the first parameter, not the return value.  Gotcha'.

The code worked perfectly fine for me on 5.3.4.  Note that the first parameter 
is not encapsulated in quotes --- it's because it's an internal persistent 
constant which evaluates to 0 (zero).  In fact, the code driving that function 
actually requires that each of the values be long, which is, of course, an 
integer variable.

You mentioned in the initial report that you're using 5.3.5 on Mac.  Are you 
using a snap or SVN checkout, or did you download the final release?

Either way, if it's not working on your 5.3.5, that may indicate that it's a bug 
in PHP itself.  What is the output of the following code?

<?php
$num = cal_days_in_month(CAL_GREGORIAN,1,2011);
echo 'CAL_GREGORIAN shows '.$num.' days this month.'.PHP_EOL;

$num = cal_days_in_month(0,2,2011);
echo 'Using zero, there are '.$num.' days next month.'.PHP_EOL;
?>

No response.