Why does Perl print a value I don't expect after incrementing?

Question

I'm running this one-liner from the command line:

perl -MList::Util=sum -E 'my $x = 0; say sum(++$x, ++$x)'

Why does it say "4" instead of "3"?

Answer 1

First, keep in mind that Perl passes by reference. That means

sum(++$x, ++$x)

is basically the same as

do {
   local @_;
   alias $_[0] = ++$x;
   alias $_[1] = ++$x;
   ∑
}

Pre-increment returns the variable itself as opposed to a copy of it*, so that means both $_[0] and $_[1] are aliased to $x. Therefore, sum sees the current value of $x (2) for both arguments.

Rule of thumb: Don't modify and read a value in the same statement.

* — This isn't documented, but you're asking why Perl is behaving the way it does.

Answer 2

You are modifying $x twice in the same statement. According to the docs, Perl will not guarantee what the result of this statements is. So it may quite be "2" or "0".