In perl 5 all order of evaluation is undefined

Not true. There are a few cases for which order of execution is very well defined. One is your example, assignment: the assignment always happens after the expression, the RHS is calculated. That makes sense, as it's only at that point that the value is known.

And two is the shortcircuit operators, and/or and &&/||. The right hand side isn't even calculated if the left hand side is all one needs to know to get the final result of the whole expression.

So this combination is very safe:

sub foo { return 3+$a };
$a = 1;
$a = foo($a) and print $a;
[download]

Order of evaluation is as follows:

• The value of $a is set to 1.
• foo($a) is calculated, returning 4.
• The new value is assigned to $a.
• and checks to see if the result of the assignment is true. It is.
• The print is executed, printing 4.

Note that this is different from the following case:

$\ = "\n";
my $x = 12;
{
  my $x = $x + 1;    # <--- point of interest
  print $x;
}
print $x;
[download]

Note that in the relevant statement, $x in the left hand side and in the right hand side are actually two different variables, both named $x. The RHS uses the outer scoped variable, the LHS the inner scoped variable. And even though here:

my $x = 123;
{
  my $x = 45 and print $x;
}
[download]

the assignment does happen before the print, it prints the outer scoped variable, because that's the variable the parameter refers to.

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