You don't need the do if you don't use the value of the last expression.

True. But it's habitual.

Moreover, I tried both ways on a 2GB file and didn't notice any difference in memory consumption:

I'm not familiar with the ins and outs of memory measurement on *nix; but I can demonstrate the difference on Windows.

C:\test>p1
[0]{} Perl> print mem;;
9,432 K

[]{} Perl> open I, '<', 'test.dat'; my $s; do{ local $/; $s = <I> }; p
+rint mem;;
997,784 K

[]{} Perl> Terminating on signal SIGINT(2)

C:\test>p1
[0]{} Perl> print mem;;
9,440 K

[]{} Perl> open I, '<', 'test.dat'; my $s = do{ local $/; <I> }; print
+ mem;;
1,986,060 K

[]{} Perl> Terminating on signal SIGINT(2)
[download]

As you can see, in the latter case, the memory assigned to the process is double: ( 997784 - 9432 ) * 2 + 9440 = 1986146; almost exactly the 1,986,060 K measured in the latter case.

The reason is that in the later case, the data is read into an internal mortal temporary scalar; and then copied from there to the named lexical, before the memory attached to the temp is freed. As the allocation is greater than (from memory) 1MB, (on windows at least) such huge allocations are allocated directly from the OS's virtual memory rather than from the process' heap; and then get released directly back to OS.

Which is usually a good thing, but it still means you need to have double the memory available for a short while, and if you are close to the limits, that can blow the process.

Is it possible that large allocations are also freed back to the OS on *nix, and you are measuring after it has been freed?