foreach has to create a fresh variable every time. Otherwise, when you keep a reference to one (as with the anonymous subroutine), they would all hold the same value every time a new value was assigned in the loop.

(And thanks to Animator for clearing up my sloppy wording on this)

Actually that's not entirely correct.

foreach does not create a variable. It is my that creates the variable. And my will use some free memory. My guess is that the references counter is increased by using the variable in an anonymous, subroutine (/closure). Which means that at the end of the foreach-block the reference counter is not 0 (and therfor the memory is not freed). Then that same part of the memory is not free at the start of the next iteration, so my will use some other memory.

Some code:
for my $x (0 .. 3) { print \$x }; ==> this code should print the same references for all 4 iterations.
for my $x (0 .. 3) { print \$x; push @z, sub { $x; }; } ==>This code will print 4 different references.

Which means Ovid’s assertion is correct. A new variable gets allocated through each iteration – only that most of the time, the space for the iterator variable from the last iteration happens to get recycled.

Makeshifts last the longest.