The first time you evaluate ($a & $b), the two arguments are strings ("12" and "8"), so it is the string version of the "&" operator. The result is 0.

But then you do a numerical comparison with $b. So $b's internal scalar caches $b's integer value along with its string value.

Now the second time you evaluate ($a & $b), $b has an integer value inside, so it is treated as an integer (this must the preference of bitwise operators?). By the table below, you can see that as long as one of the arguments to the "&" operator is considered to be an integer, we use the integer/bitwise version of the operator (and the result is 8 this time).

print "12" & "8", $/;  # 0
print  12  &  8,  $/;  # 8
print  12  & "8", $/;  # 8
print "12" &  8,  $/;  # 8
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