Re^5: $1 not "freezing" in an addition

[grondilu]

in reply to Re^4: $1 not "freezing" in an addition
in thread $1 not "freezing" in an addition

Interesting. This gave me the idea of trying this code:

$_ = 1; print $_ + $_++;

Amazingly enough, this prints 3, not 2

But this does print 2:

$_ = 1; print 0+$_ + $_++;

[tobyink]

I can count as high as 7...

sub r  { $_[0] } # no-op?
sub rr :lvalue { $_[0] } # no-op?

$_ = 1;
print ($_) + $_++;

$_ = 1;
print r($_) + $_++;

$_ = 1;
print $_ + $_++;

$_ = 1;
print $_ + ++$_;

$_ = 1;
print r(++$_) + r(++$_);

$_ = 1;
print ++$_ + ++$_;

$_ = 1;
print rr(++$_) + rr(++$_)++;

(Requires Perl 5.16, though you can get up to 6 in Perl 5.10 and above.)

Update: 0..10

use v5.16;
no v5.17;
no strict 'subs';

sub r          { $_[0] } # no-op?
sub rr :lvalue { $_[0] } # no-op?

$_ = 1;
say ( $ ++ _ ) + $_++;

$_ = 1;
say ($_) + $_++;

$_ = 1;
say r($_) + $_++;

$_ = 1;
say $_ + $_++;

$_ = 1;
say $_ + ++$_;

$_ = 1;
say r(++$_) + r(++$_);

$_ = 1;
say ++$_ + ++$_;

$_ = 1;
say rr(++$_) + rr(++$_)++;

$_ = 1;
say ++rr(++$_) + rr(++$_);

$_ = 1;
say ++rr(++$_) + rr(++$_)++;

$_ = 1;
say ++rr(++$_) + ++rr(r(rr(++$_)++));