Hey Hugues,

both of them are scalar context, but in the first you wanna place a hash to the scalar variable that is why you have the error, and in the second you assign a hash reference to the scalar, which is ok, because the reference is a scalar.

This is a hash:

%hash1 = (
        key1 => 'value1',
        key2 => 'value2',
        key3 => 'value3',
    );

You cannot store a hash in a scalar, but you can store the reference to the hash in a scalar:

my $hash_ref1 = \%hash;

But you don't wanna deal with a hash name because you would use it only once to create the hash, so you skip this step and create an anonymous hash, and place the reference to a scalar in one step:

my $hash_ref2 =   {
        key1 => 'value1',
        key2 => 'value2',
        key3 => 'value3',
    };

You can read more about it in the Intermediate Perl book from O'Reilly. 

Best regards,
Attila


On Tue, Apr 22, 2014 at 5:31 PM, Hugues <hugues@max4mail.com> wrote:
Thanks for your help
if I understand well

$Stock->{$NumP} = ( $Num =>  { Code    => $parc->{Code} .... )  is scalar context, and generate a error
$Stock->{$NumP} = { $Num =>  { Code    => $parc->{Code} .... } is correct.

I will add "use diagnostics;" in my code
thanks
Hugues

Le 22/04/2014 17:18, Maxwell Carey a écrit :

On 04/22/2014 08:21 AM, Hugues wrote:
$Stock->{$NumP} = ( $Num =>  { Code    => $parc->{Code},
NumP => $parc->{NumP},
NumS => $parc->{NumS},
NumC => $parc->{NumC}
                                                            } );

Not related to Dancer, just plain Perl. You're assigning a list to a scalar, which is probably not what you intend. What you're doing is the same as:

    my $foo = ( "foo", "bar" );

which sets the value of $foo to "bar". If you turn on diagnostics with `use diagnostics;`, you can get a nice explanation of the warning:

    Another common error is to use ordinary parentheses to construct a list
    reference when you should be using square or curly brackets, for
    example, if you say

        $array = (1,2);

    when you should have said

        $array = [1,2];

    The square brackets explicitly turn a list value into a scalar value,
    while parentheses do not.  So when a parenthesized list is evaluated in
    a scalar context, the comma is treated like C's comma operator, which
    throws away the left argument, which is not what you want. See
    perlref for more on this.

To fix, change the parentheses to curly braces to create an anonymous hash:

    $Stock->{$NumP} = { $Num =>  { Code    => $parc->{Code},
NumP => $parc->{NumP},
NumS => $parc->{NumS},
NumC => $parc->{NumC}
                                                            } };

Or simply remove the additional level of hash nesting (which is essentially what is happening in your current program):

    $Stock->{$NumP} = { Code    => $parc->{Code},
                                          NumP => $parc->{NumP},
                                          NumS => $parc->{NumS},
                                          NumC => $parc->{NumC}
                                       };
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