Re: Re: How's your Perl?

[davido]

Though I'm almost certain that's what the OP had in mind, it relies on a bug rather than a documented feaure. For that reason I happen to like this solution to question #5 of the OP's quiz:

sub{my$x=\$x};

Updated version that creates only one $x:

sub { my $x; $x = \$x; }

The anonymous sub is created and immediately falls out of existance because it's not passed to a scalar variable. And yet $x never disappears because its reference count is always going to be 1; it refers to itself.

It probably doesn't qualify as a static, and is pretty much useless, but it meets the definition of static that the OP gave.

[Anonymous Monk]

I do not think your code does what you think it does. You have two different $x variables shown in your anonymous subroutine, not a single one referencing itself.

[Anonymous Monk]

A static variable in a subroutine context would be created only once and hold its value between invocations. You've created a circular reference in an uninvokable routine. Even if it weren't optimized away, of which I am uncertain, it bears no resemblance to a static variable.

[davido]

You are correct, given the conventional definition of a static variable. But the Original Poster provided us with a test of expert Perl prowess, and went so far as to give us his expert definition of static:

"static variable" means a variable with infinite lifetime...

My solution meets his spec. In a rediculous way, of course... but then again it's a rediculous quiz.

[xmath]

But the Original Poster provided us with a test of expert Perl prowess, and went so far as to give us his expert definition of static

ok ok I messed up; because I've had questions about what a "static variable" is (not all perl programmers are C programmers) so I added a quick clarification and got it wrong.. nobody's perfect

no need to get sarcastic here

My solution meets his spec. In a rediculous way, of course... but then again it's a rediculous quiz.

true on both points   :-)

[xmath]

oops, you're right, my definition of "static" in the clarification was flawed :-)

I do mean function-scoped static like in C/C++ ofcourse

btw tachyon, your #1 solution isn't:
perl -le 'my @x = ( \$_, \$_ ); $x[0]=42; print $x[1]'

[tachyon]

Read your rules. You don't specify HOW the value of $x[0] is changed :-) vis Create an array @x such that changing $x[0] also sets $x[1] to the same value While assignment is one way to change a value....

@x=(\$_,\$_);
$_=10;
print "\$x[$_] = ${$x[$_]}\n" for 0..$#x;
$_=20;
print "\$x[$_] = ${$x[$_]}\n" for 0..$#x;
__DATA__
$x[0] = 10
$x[1] = 10
$x[0] = 20
$x[1] = 20

[Juerd]