There are many factors to consider when comparing the performance of your two models.

First, substr() and unpack() behave differently:

• substr() copies the sub-string.
• unpack('A30') copies the sub-string, scans it to locate the end of the space padded string, and shrinks the sub-string such that it does not include the padding.

Secondly, Perl treats scalar assignments different from list assignments. In fact, the scalar assignment op-code is implemented much simpler. Benchmark the difference between "($a) = (1)" and "$a = 1" to see the effect that this has. Although 15+ scalar assignments may be more expensive than a list assignment that involves 15+ values, the overhead of a list assignment dampens the overhead of having many scalar assignments.

The only true way of comparing the two approaches, is the method that you have chosen -- benchmarking.

If you need to improve the performance of your substr() model, one (ugly) approach may be to use one large list assignment ("@data = (substr(), substr(), ...)") the same as you do for unpack(). Again, only benchmarking can tell... :-)

UPDATE: substr() only uses magic if called in lvalue context.