Re^2: method dispatch question

Indeed, I missed it, and it works very well. But now let us make the problem a bit more complicated (I see that I have simplified to much from my original problem): Assume that $mref does not come in as parameter, but will be calculated somehow; for example:

my $mref=$self->get_mref(...); # Calculate $mref
$self->$mref(...); # use it to dispatch
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Can this still be written in one line, without intermediate variable $mref? I learned that my original id

$self->&{ $self->get_mref(...) }()
[download]

does not work, but extending your suggestion to this case would, I think, fail too:

$self->($self->get_mref(...))()
[download]

Can this (now more complicated) example be easily solved too?

--
Ronald Fischer <ynnor@mm.st>

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Re^3: method dispatch question by moritz (Cardinal) on Jul 09, 2008 at 12:20 UTC
Even if there is a solution for that (and I don't doubt there is one, perl is full of surprising syntax ;-), you shouldn't go for it, because it's not really readable anymore. Use a temporary variable instead. `$self->($self->get_mref(...))()` won't work because `$something->(...)` treats `$something` as a sub ref. If there's a way, I suspect it involves `${...}`.	[reply] [d/l] [select]
Re^4: method dispatch question by rhesa (Vicar) on Jul 09, 2008 at 12:41 UTC
I think `$self->${ \ $self->get_mref(...) }(...)` would work, but I agree with you about the readability.	[reply] [d/l]
Re^3: method dispatch question by rafl (Friar) on Jul 09, 2008 at 12:35 UTC
Can this still be written in one line, without intermediate variable $mref? `$self->get_mref->($self, @args);` [download] or make get_mref return a code reference that captured $self so you can omit it when calling it. `sub get_mref { my ($self) = @_; my $coderef = \&some_method; return sub { $self->$coderef(@_) }; } $self->get_mref->(@args);` [download]	[reply] [d/l] [select]
Re^3: method dispatch question by Arunbear (Prior) on Jul 09, 2008 at 12:34 UTC
There is a way: `use strict; use CGI; sub meth {'p'} my $cgi = CGI->new; print $cgi->${\meth()}('foo')."\n";` [download] but why would you want to do that? update D'oh! didn't read the question closely enough. This is not what you're looking for.	[reply] [d/l]