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Slice returned hash or access to its elements without assigning to a variable
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| Andrew 2005-06-11, 3:57 pm |
| Hi,
Please help me in solving next problem(I search in FAQ bun have not
find an answer):
There is a sub foo which return hash:
sub foo{my %in=( key1=>value1
key2=>value2
key3=>value3);
}
I need to assess elements of returned hash without assigning to a
variable. Something like this:
print @&foo(){'key1','key2'};
print $&foo(){'key1'};
But this does not work: "Bareword found where operator expected"
Thanks,
Andrew
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| Glenn Jackman 2005-06-11, 3:57 pm |
| At 2005-06-11 12:35PM, Andrew <les_andrey@yahoo.com> wrote:
> Hi,
>
> Please help me in solving next problem(I search in FAQ bun have not
> find an answer):
> There is a sub foo which return hash:
>
> sub foo{my %in=( key1=>value1
> key2=>value2
> key3=>value3);
> }
Strictly speaking, this returns a 6-element list, not a hash.
> I need to assess elements of returned hash without assigning to a
> variable. Something like this:
> print @&foo(){'key1','key2'};
> print $&foo(){'key1'};
>
> But this does not work: "Bareword found where operator expected"
Coerce the returned list from foo into a hash ref, and dereference
accordingly:
sub foo { (a=>1,b=>2,c=>3) }
$b = +{foo()}->{b};
@arr = @{%{+{foo()}}}{'a','b'}
--
Glenn Jackman
NCF Sy min
glennj@ncf.ca
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| Glenn Jackman 2005-06-11, 3:57 pm |
| At 2005-06-11 12:35PM, Andrew <les_andrey@yahoo.com> wrote:
> Hi,
>
> Please help me in solving next problem(I search in FAQ bun have not
> find an answer):
> There is a sub foo which return hash:
>
> sub foo{my %in=( key1=>value1
> key2=>value2
> key3=>value3);
> }
Strictly speaking, this would return a 6-element list, not a hash, if it
was proper Perl.
sub foo { (a=>1,b=>2,c=>3) }
> I need to assess elements of returned hash without assigning to a
> variable. Something like this:
> print @&foo(){'key1','key2'};
> print $&foo(){'key1'};
>
> But this does not work: "Bareword found where operator expected"
Coerce the returned list from foo into a hash ref, and dereference
accordingly:
$b = +{foo()}->{b};
@arr = @{%{+{foo()}}}{'a','b'}
--
Glenn Jackman
NCF Symin
glennj@ncf.ca
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