my @arr = (
[1,2],
[3,4],
);
my %hash = ( FirstKey => \@arr );

my $aref = $hash{FirstKey};
my $row0 = $aref->[0];
push @$row0, 99;

use Data::Dumper;
print Dumper \%hash;
[download]

Further to runrig's replies:
It's important to understand at each step how the structures of the hash and arrays are changing and evolving. Here's a kind of "running commentary" on that evolution, and a possible way to access and print what you want. Liberal use of data structure (for that's what you're dealing with; see perldsc) dump statements can be very enlightening about the changes that are going on. (And BTW, congratulations on being able to work with a system having at least 128 Terabytes of system RAM — or do I misinterpret the reference address of 0x7ffafc013bf8?)

>perl -wMstrict -MData::Dump -le
"my %mapp;
 my @arr = qw(R T HH M);
 ;;
 $mapp{k} = \@arr;
 my @val = $mapp{k};
 ;;
 dd \%mapp;
 dd \@val;
 ;;
 print join q{,}, @{$mapp{k}};
 ;;
 push @val, 'RBB';
 dd \@val;
 ;;
 $mapp{k} = \@val;
 dd \%mapp;
 ;;
 print join q{,}, @{$mapp{k}};
 print join q{,}, @{ $mapp{k}->[0] }, $mapp{k}->[1];
"
{ k => ["R", "T", "HH", "M"] }

[["R", "T", "HH", "M"]]

R,T,HH,M
[["R", "T", "HH", "M"], "RBB"]

{ k => [["R", "T", "HH", "M"], "RBB"] }

ARRAY(0x1e14b3c),RBB
R,T,HH,M,RBB
[download]

(And BTW, congratulations on being able to work with a system having at least 128 Terabytes of system RAM — or do I misinterpret the reference address of 0x7ffafc013bf8?)

With virtual memory, heap does not have to be contiguous.

This C program allocates a single page of memory starting at the 4GB base address and then doubles the base address each time until it fails:

#define  _CRT_SECURE_NO_WARNINGS
#include <windows.h>
#include <conio.h>
#include <time.h>
#include "debug.h"
#include "mytypes.h"

int main( int argc, char **argv ) {
    U64 size = 4096, base = 4294967296;
    char *p;

    while( 1 ) {
        char *p = VirtualAllocEx(
            GetCurrentProcess(),
            (VOID*)base, size,
            MEM_RESERVE | MEM_COMMIT, PAGE_READWRITE
        );
        DIEIF( p == NULL );
        printf( "4k allocated at:%16p (%30I64d)\n", p, 0+p );
        _getch();
        base *= 2;
    }
    exit( 0 );
}
[download]

I only have 8GB of ram, but the last time the above succeeds:

C:\test>vmem3
4k allocated at:0000000100000000 (                    4294967296)
4k allocated at:0000000200000000 (                    8589934592)
4k allocated at:0000000400000000 (                   17179869184)
4k allocated at:0000000800000000 (                   34359738368)
4k allocated at:0000001000000000 (                   68719476736)
4k allocated at:0000002000000000 (                  137438953472)
4k allocated at:0000004000000000 (                  274877906944)
4k allocated at:0000008000000000 (                  549755813888)
4k allocated at:0000010000000000 (                 1099511627776)
4k allocated at:0000020000000000 (                 2199023255552)
4k allocated at:0000040000000000 (                 4398046511104)
[ 1916] vmem3.c(18):p == NULL
        failed with error 87 (The parameter is incorrect)
[download]

It allocated a page at a base address that is 4096 GB. At that point the process is only using 668k of memory.

4096GB is 2^42 which reflects that my 64-bit processor actually only has a 44-bit virtual address space.

---

With the rise and rise of 'Social' network sites: 'Computers are making people easier to use everyday'

Examine what is said, not who speaks -- Silence betokens consent -- Love the truth but pardon error.

"Science is about questioning the status quo. Questioning authority".

In the absence of evidence, opinion is indistinguishable from prejudice.