In short, "don't do that". But if you're interested in the internal details, read on.

Values in Perl are stored in a struct call "SV" (or a derivative thereof). SVs aren't copied around. Pointers to SVs are. That results in "pass by reference" being the default model for every operator, function and sub. This creates interesting side-effects as seen in the following:

>perl -le"$x=3; print($x+0, $x, ++$x, $x, $x++, $x, $x+0);"
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$x and ++$x place a pointer to $x's SV on the stack. $x+0 (and "$x" for strings) creates a new SV, so it's no affected by later changes to $x. Similarly, $x++ creates a new SV containing a copy of what $x was at the time of the increment.

#include <stdio.h>

void print(const int &g, const int &f, const int &e, const int &d, con
+st int &c, const int &b, const int &a) {
   printf("%d%d%d%d%d%d%d\n", a, b, c, d, e, f, g);
}

int main() {
   int x = 3;
   print(0+x, x, x++, x, ++x, x, 0+x);
   //    5    5  4    5  5    5  3
   return 0;
}
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