In order for the code to be truly equivalent the s modifier should be used on the substitution or a newline may break it.

$_ = "  foo\nbar  ";

s{ ^\s* (\S.*?) \s*$ }{$1}x;

print "<$_>";

__END__
<  foo
bar  >
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I assume you added the \S in the pattern as an improvement, but it should perhaps be noted that it has the effect of leaving a line of only whitespaces untouched, whereas the other ways don't.

lodin