On 29/01/2013 11:36, gvim wrote:
On 29/01/2013 02:33, Mike Stok wrote:
Have you tried reading
perldoc -f scalar
Hope this helps,
Mike
Yes, I'm aware of the scalar function but still not clear why
assigning $r->method as a hash value doesn't invoke a scalar context
in the first place.
The code
return { passed => $passed, valid => $r->valid, missing => $r->missing, invalid =>
$r->invalid, unknown => $r->unknown };
is equivalent to this
return { 'passed', $passed, 'valid', $r->valid, 'missing', $r->missing, 'invalid',
$r->invalid, 'unknown', $r->unknown };
The bit inside the curly braces is not any kind of magical "hash
constructor" but a list, plain and simple.
None of the values in that list "know" that they will eventually become
a hash key or a hash value.
Everything in there is just in list context. So your method calls are in
list context.
The curly braces then take the elements of that list, in pairs, and then
treat each pair as (key, value) to build a hash.
(This links to a common discussion (linked in turn with the old 'return
undef;' vs. 'return;' argument) about whether or not functions should
behave differently depending on scalar and list context.)
Regards,
Bill
