Arguments are passed to subs as a list. if you have foo($arg1,@moargs) it will recieve @_ = ($arg1,@moargs). In list context list elements are placed in the list in order. However prototypes can change the way the arguments are evaluated. foo (\@\%) causes perl to treat foo(@argar,%arghs) as foo(\@argar,\%arghs).

This is all explained in perlsub which draws on information in perlfunc and perlop.

In reply to Re: eval order of args to a sub by Trizor
in thread eval order of args to a sub by otto

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