the $?>>8 construct shifts the $? value to the right by 8 bits. In order to examine the entire value, don't bit-shift it. However, perldoc perlvar has this to say about $?:

This is just the 16-bit status word returned by the wait() system call (or else is made up to look like it).

Where I come from, 16-bit values equate to 2-byte values, not four.