perldoc perlref
Making References
References can be created in several ways.
1. By using the backslash operator on a variable, subroutine, or
value. (This works much like the & (address-of) opera-
tor in C.) This typically creates another reference to a
variable, because there’s already a reference to the vari-
able in the symbol table. But the symbol table reference might
go away, and you’ll still have the reference that the
backslash returned. Here are some examples:
$scalarref = \$foo;
$arrayref = \...@argv;
$hashref = \%ENV;
$coderef = \&handler;
$globref = \*foo;
It isn’t possible to create a true reference to an IO handle
(filehandle or dirhandle) using the backslash operator.
The most you can get is a reference to a typeglob, which is
actually a complete symbol table entry. But see the
explanation of the *foo{THING} syntax below. However, you can
still use type globs and globrefs as though they were
IO handles.
2. A reference to an anonymous array can be created using square
brackets:
$arrayref = [1, 2, [’a’, ’b’, ’c’]];
Here we’ve created a reference to an anonymous array of three
elements whose final element is itself a reference to
another anonymous array of three elements. (The multidimensional
syntax described later can be used to access this.
For example, after the above, "$arrayref->[2][1]" would have the
value "b".)
///////// 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！
@list = (\$a, \...@b, \%c);
@list = \($a, @b, %c); # same thing!
As a special case, "\(@foo)" returns a list of references to the
contents of @foo, not a reference to @foo itself.
Likewise for %foo, except that the key references are to copies
(since the keys are just strings rather than full-
fledged scalars).
> -----邮件原件-----
> 发件人: Mark_Galeck [mailto:mark_galeck_spam_mag...@yahoo.com]
> 发送时间: 2009年11月日 11:53
> 收件人: beginners@perl.org
> 主题: reference to anonymous array, references last element instead??
>
> Why does
>
> $foobar = \("foo", "bar");
> print $$foobar;
>
>
> print "bar" ??
>
> Thank you for any insight. Mark
>
>
> --
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