Ireneusz Pluta <ipl...@wp.pl> writes:
> Harry Putnam wrote:
>> #!/usr/local/bin/perl
>>
>> use strict;
>> use warnings;
>>
>> my $var1 = 'whoopdee';
>> my $var2 = 'do';
>>
>> my %dispatch = (
>> y => \&yy($var1,$var2),
> this, actually, is not a code reference but a return value reference
> of the &yy($var1, $var2) subroutine call, executed right at the time
> of %dispatch assignment. By using (...) after a subroutine name you
> have the subroutine executed and get its return value. Then, the
> leading '\' in this syntax makes a reference to that return value.
Thanks. I'm not sure I really understand, why is the subroutine NOT called
immediately if (..) is not present?
I guess what I'm asking is how is it that:
y => \&yy,
Is not executed but:
y => \&yy($var,$var),
is executed?
If you remove the reference notation in other code like this:
my $it = &yy;
my $it = yy($var,$var);
Either one would execute immediately right?
> To see what has happened, try to run your code in debug mode (perl -d)
> and view the contents of your %dispatch hash, or place
>
> use Data::Dumper;
> print Dumper { %dispatch };
Thats a good tip... thanks. I recall having seen this mentioned
before but I didn't understand what it meant.
I see now how you can tell that \&yy($var,$var) executed immediately.
[...]
> 'y' => \'You pressed `y\',whoopdee do .. but why?
[...]
>> my $code = $dispatch{$selection} || $dispatch{'error'} ;
>> $code->();
> this is the place you really want to call your subroutine, so also the
> place to pass your arguments to it:
>
> $code->($var1, $var2);
>> }
Ok, thanks, I hadn't understood how that notation `$code->()' worked.
I started by copying a googled example of a dipatch table and tried to
edit it for my own use, but I have A LOT of trouble following along in
code... it seems some of the posters here read it like a book.
Hopefully I will be able to eventually too.
However, using your suggestion:
> $code->($var1, $var2);
and the Dumper lines:
use Data::Dumper;
print Dumper { %dispatch };
I don't see the expected result when I press `y'. (The code is at the end)
It seems to do nothing.
The dump shows:
$VAR1 = {
'y' => sub { "DUMMY" },
'n' => sub { "DUMMY" },
'q' => sub { "DUMMY" },
'error' => sub { "DUMMY" }
};
I guess it just shows that nothing happens...
But the code is at the end... no doubt riddled with more mistakes.
>> sub yy {
>> my ($var1,$var2);
>> ($var1,$var2) = (shift,shift);
> why not:
>
> my ($var1, $var2) = @_;
I guess I wasn't sure that would get the right result but knew
($var1,$var2) = (shift,shift); would. Your example is simpler
and more obvious. Thanks
> my $var1 = shift;
> my $var2 = shift;
Is that actually better in some way? Or do you just mean it looks
nicer?
------- --------- ---=--- --------- --------
non working code:
#!/usr/local/bin/perl
use strict;
use warnings;
my $var1 = 'whoopdee';
my $var2 = 'do';
my %dispatch = (
y => \&yy,
n => sub { print "You pressed \`n' \n";},
q => sub { print "Goodbye\n" and exit;},
error => sub { print "invalid selection\n" }
);
use Data::Dumper;
print Dumper { %dispatch };
while(1)
{
print "press y \n",
"press n \n",
"press q to Exit\n";
chomp(my $selection = <STDIN>);
my $code = $dispatch{$selection} || $dispatch{'error'} ;
$code->($var1,$var2);
}
sub yy {
my ($var1,$var2);
($var1,$var2) = @_;
my $retstr = sprintf "You pressed \`y',$var1 $var2 .. but why?\n";
return $retstr;
}
------- --------- ---=--- --------- --------
output from running: ./dispatch (the yy() function doesn't do anything.
$VAR1 = {
'y' => sub { "DUMMY" },
'n' => sub { "DUMMY" },
'q' => sub { "DUMMY" },
'error' => sub { "DUMMY" }
};
press y
press n
press q to Exit
y
press y
press n
press q to Exit
n
You pressed `n'
press y
press n
press q to Exit
q
Goodbye
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