I think I made an interesting discovery. As you suggest above, if I do this it only makes one copy of the file in memory:

P:\test>perl -le"
my $s; do{ local $/; $s=<>}; 
print length $s; 
system qq[tasklist /fi \"pid eq $$\" ]" 1000000.dat

11000001

Image Name                   PID Session Name     Session#    Mem Usag
+e
========================= ====== ================ ======== ===========
+=
perl.exe                     604                         0     12,860 
+K
[download]

And if I do it this way it uses two copies:

P:\test>perl -le"
my $s=do{ local $/; <>}; 
print length $s; 
system qq[tasklist /fi \"pid eq $$\" ]" 1000000.dat

11000001

Image Name                   PID Session Name     Session#    Mem Usag
+e
========================= ====== ================ ======== ===========
+=
perl.exe                    1372                         0     23,624 
+K
[download]

But what fooled me, as I was using my perl shell, is that if you eval the statement, only one copy is made:

P:\test>perl -le"
my $s = eval q[do{ local $/; <>}]; 
print length $s; 
system qq[tasklist /fi \"pid eq $$\" ]" 1000000.dat

11000001

Image Name                   PID Session Name     Session#    Mem Usag
+e
========================= ====== ================ ======== ===========
+=
perl.exe                    1764                         0     12,880 
+K
[download]

And I can't quite figure out why that should be so?