From perldoc perlsub in the first paragraph under prototypes:

The declaration of the function to be called must be visible at compile time. The prototype affects only the interpretation of new-style calls to the function, where new-style is defined as not using the & character. In other words, if you call it like a builtin function, then it behaves like a builtin function. If you call it like an old-fashioned subroutine, then it behaves like an old-fashioned subroutine. It naturally falls out from this rule that prototypes have no influence on subroutine references like \&foo or on indirect subroutine calls like &{$subref} or $subref->().

I think that pretty much explains it.