Perl evaluates the arguments from left to right (according to the associtivity of the list seperator).

>perl -le"$i=3; print($i+0, ++$i, $i+0);"
344
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However, because Perl passes all arguments by reference, the order of operation sometimes *appears* unclear.

>perl -le"$i=3; print($i, ++$i, $i);"
444
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The first snippet is similar to

# $i=3; print($i+0, ++$i, $i+0);
$i = 3;
do {
   local @_;
   $anon0 = $i+0;  alias $_[0] = $anon0;
   $i = $i+1;      alias $_[1] = $i;
   $anon1 = $i+0;  alias $_[2] = $anon1;
   &print
};
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The second snippet is similar to

# $i=3; print($i, ++$i, $i);
$i = 3;
do {
   local @_;
   $i;         alias $_[0] = $i;
   $i = $i+1;  alias $_[1] = $i;
   $i;         alias $_[2] = $i;
   &print
};
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Can you guess what the following prints?

perl -le"$i=3; sub { $_[1]++; print @_ }->($i+0, ++$i, $i+0, $i);"
[download]