in reply to Re^6: Adding Unique Elements to Array
in thread Adding Unique Elements to Array

No, the behavior of assignment and the x operator are defined for array slices.

Both () x EXPR and () = EXPR are defined, but () x= EXPR is another operator and doesn't work like you'd expect; in fact, () x= isn't documented and x is the only operator that can be used with = that has special semantics for its LHS operand. All other force scalar context so it's a no-brainer how they'll work for list assignment. perlop says that it works like in C, and you can then assume the following paragraph is about scalar assignment and it always talks about a singular lvalue. Only in the last paragraph list assignment is covered.

In (context()) = EXPR the subroutine &context is called in list context; in (context()) x= EXPR it's called in scalar context. However, the lvalue of all elements in the list in (LIST) x= EXPR are taken (hence the autovivification), just as for

sub foo {} foo(@foo{qw/a b/}); print keys %foo; # ab
The lvalues are then put in scalar context making them not a list anymore but a sequence of scalars with the scalar-comma between them (see perlop). Why this happens I have no idea. As a consequence of the "list" of elements being "comma-separated in scalar context" all values but the last are put in void context and the last in scalar context. Example:
use warnings; sub context { defined $_[0] ? $_[0] ? 'list' : 'scalar' : 'void' } sub foo :lvalue { print context(wantarray); my $dummy = '' } sub bar :lvalue { print context(wantarray); my $dummy = '' } my $baz; ($baz, foo(), bar()) x= 3; __END__ Useless use of private variable in void context void scalar

Is \ documented for array slices?

From perlref

Taking a reference to an enumerated list is not the same as using square brackets--instead it's the same as creating a list of references!

The explicit parenthesis isn't needed. (Just consider \qw/ foo bar /.) @foo[@bar] is a listop called aslice. I don't know if it's documented that aslice may be used instead of a list, so you may be right. It's at least more documented than the other tricks. ;-)

ihb

See perltoc if you don't know which perldoc to read!

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Re^8: Adding Unique Elements to Array
by Roy Johnson (Monsignor) on Mar 01, 2005 at 21:32 UTC
    You are right. This gets a compilation error:
    my @foo = qw(one two three); @foo x= 3;
    I had thought that all combination operators were defined in terms of their expansion: if you can write @foo = @foo x 3, you can write @foo x= 3. Not so. Good to know.

    Caution: Contents may have been coded under pressure.