Reference to a function return

[yoxoman]

Hello,

In the expressions

my $ref = \foo();

or

my $ref = \$myobj->foo

$ref is a reference to a scalar, even if the foo function returns an
array (in that case, $ref points to an element of array...)

Do you know why ?


Thanks.

 

[John Bokma]

Hello,

In the expressions

my $ref = \foo();

or

my $ref = \$myobj->foo

$ref is a reference to a scalar, even if the foo function returns an
array (in that case, $ref points to an element of array...)

Do you know why ?

I guess you want:

my $ref_to_return = [ foo() ];

 

[C.DeRykus]

Hello,

In the expressions

my $ref = \foo();

or

my $ref = \$myobj->foo

$ref is a reference to a scalar, even if the foo function returns an
array (in that case, $ref points to an element of array...)
Do you know why?

That's because the members of the array are
returned as a list. The LHS is a scalar so
the comma operator acts on that list and the
last member of the list gets assigned to the
scalar.

See: perldoc -q list
"What is the difference between a list
and an array?" ...

Since \ provides a list, you might want to
just use an array on the LHS:

my @refs = \$myobj->foo.