Re: Perl Idioms Explained - ${\$obj->method} and @{[sort @list]}

[Abigail-II]

in reply to Perl Idioms Explained - ${\$obj->method} and @{[sort @list]}

One must be careful with the ${\foo} because if foo returns something that isn't a simple scalar value then you might get unexpected results.

To be more precise, the \ operator gives its argument list context, which may surprise many people (it did me). The other interpolation construct @{[  ]} also provides list context. This is something to be aware of, and use of scalar will be appropriate in some cases, including in ${\scalar foo}.

Abigail

[bradb]

To be more precise, the \ operator gives its argument list context, which may surprise many people (it did me).

Why would this surprise you? :)

If I ask for a reference to an array that contains ('foo', 'bar', 'baz'), I expect a reference to an array that has three elements in it, not the number 3 (as would be the case if \ provided scalar context for its arguments.)

[Aristotle]

That doesn't explain it. \('foo', 'bar', 'baz') is the same as (\'foo', \'bar', \'baz'). Taking a reference to an array certainly doesn't behave that way.

Makeshifts last the longest.

[adrianh]

Read Arrays are not lists and Scalars, Lists, and Arrays and repeat to yourself three times - "('foo', 'bar', 'baz') is not an array" ;-)

I'm with Abigail-II. The \ operator has lots of weird cases that surprise you. For example:

\@foo       # reference to array @foo
\(@foo, $a) # list of reference to array @foo and scalar $a
\(@foo)     # list of references to contents of @foo

[bradb]

Read Arrays are not lists and Scalars, Lists, and Arrays and repeat to yourself three times - "('foo', 'bar', 'baz') is not an array" ;-)

I can't be blamed if you don't read my post. ;-) I didn't say ('foo', 'bar', 'baz') was an array, I said If I ask for a reference to an array that contains ('foo', 'bar', 'baz').

HTH.