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

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

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

开发者_高级运维

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

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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.

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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".

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Because both incrementers are executed before the sum is calculated.

After both execute, x = 2.

2 + 2 = 4.