$/ is just a newline by default. It's there to break the line and set the next shell prompt back to the far left of the screen.

On 32-bit perl, 

$ perl -e'print 1<<$_, $/ for 28..36'
268435456
536870912
1073741824
2147483648
1
2
4
8
16
$

[download]
Perl 1<<$n is equivalent to 2**$n, not 2**($n+1). The << operator is regarded as bitwise by perl, so forces the result into a bitfield, an unsigned int. As in C, an overflowed int wraps to its low bits[What was I thinking there?]. On 64-bit perl, the above gives, 
$ perl -e'print 1<<$_, $/ for 28..36'
268435456
536870912
1073741824
2147483648
4294967296
8589934592
17179869184
34359738368
68719476736
$

[download]
For both 32- and 64-bit Perl, 1<<64 == 1.

After Compline,
Zaxo

In reply to Re^3: 64 bit perl by Zaxo
in thread 64 bit perl by oopplz

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