Are all references live in scalars? even explicit ones?

Question

For instance:

use strict;
use warnings;
use Devel::Peek;

Dump({});

Would print the following:

SV = IV(0x170fc98) at 0x170fca8
  REFCNT = 1
  FLAGS = (TEMP,ROK)
  RV = 0x170fa68
  SV = PVHV(0x1715970) at 0x170fa68
    REFCNT = 1
    FLAGS = (SHAREKEYS)
    ARRAY = 0x0
    KEYS = 0
    FILL = 0
    MAX = 7
    RITER = -1
    EITER = 0x0

it seems like SV = IV(0x170fc98) at 0x170fca8 (a numerical, IV, scalar) references the PVHV (hash) SV = PVHV(0x1715970) at 0x170fa68

I was expecting something like:

  SV = PVHV(0x1715970) at 0x170fa68
    REFCNT = 1
    FLAGS = (SHAREKEYS)
    ARRAY = 0x0
    KEYS = 0
    FILL = 0
    MAX = 7
    RITER = -1
    EITER = 0x0

PVHV is not a hash reference. It's a hash. – ikegami Jan 02 '14 at 05:37 — ikegami, Jan 02 '14 at 05:37
@ikegami, typo, sorry. thanks for letting me know. – nixer Jan 02 '14 at 07:02 — nixer, Jan 02 '14 at 07:02

score 3 · Answer 1 · edited Jan 02 '14 at 05:34

3

Yes, references are always scalars.

The construct { ... } creates an empty hash, then returns a reference to it, which is a scalar.

It's similar to do { my %anon = ( ... ); \%anon }.

edited Jan 02 '14 at 05:34

ikegami

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answered Jan 01 '14 at 20:45

tobyink

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score 3 · Accepted Answer · answered Jan 01 '14 at 20:45

3

Each data type in Perl has a common “header” of data that supplies reference counting, etc. The flags in this header determine what type it actually is (e.g. a PVHV). So internally a %hash is a PVHV. The header is followed by other hash-specific fields.

Now what is a $reference = \%hash? A reference is a scalar containing a pointer to another data type. This is implemented as the integer field in an IV, but the ROK flag is set to show that the data isn't an integer but a pointer.

You can only pass scalars to a Perl subroutine, therefore you can only pass a hash reference, not a hash itself to Dump. As Dump prints out whatever scalar you provide, the enclosing reference is also shown.

To understand internal data types in Perl better, I recommend the Illustrated Perl Guts.

answered Jan 01 '14 at 20:45

amon

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$reference = \%hash which assign the ref of %hash to the $reference scalar seems ok to me to have the IV header, but my example which is using only {}, I was actually expecting to see only PVHV. – nixer Jan 01 '14 at 20:53
@naabu That's because `{...}` doesn't create a hash, it constructs a hash reference. It was designed that way because of back-compatibility: collections like `@arrays` or `%hashes` can only contain scalar values, and having `{}` create a reference allows us to write nested literals: `[{foo => 2}, 3, [1, 2, {3 => ["a", "b"]}]]` (`[...]` creates an array reference). – amon Jan 01 '14 at 21:12
1

things were maybe a bit clearer when there were separate IV and RV (reference value) types, but when things were reorganized (for speed and less memory, in 5.10?) they were united. – ysth Jan 01 '14 at 21:47
@naabu, The above comment should read: That's because `{...}` doesn't **just** create a hash, it **also** constructs a reference to it. – ikegami Jan 02 '14 at 05:36

Are all references live in scalars? even explicit ones?

2 Answers2