@_ is only passed unchanged to functions called using a prefixing & and no () (e.g. &func). However shift is not really a function. That means you cannot call it like &shift. Shift also takes its arguments in a special way so the & syntax would not even make sense, because & ignores prototypes (which are needed to obtain this "special" treatment of arguments from perl code.)

arturo is mostly correct about what shift defaults to, but note that there are exceptions (that you are unlikely to encounter anytime soon) where shift will take @ARGV by default within a subroutine. See the docs for more details.

In reply to Re: @_ the default variable? by wog
in thread @_ the default variable? by Anonymous Monk

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