perl - Trouble with shift and dereference operator -

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i have question regarding how left , right sides of -> operator evaluated. consider following code:

#! /usr/bin/perl  use strict; use warnings; use feature ':5.10';  $, = ': '; $" = ', ';  $sub = sub { "@_" };  sub u { shift->(@_) } sub v { $s = shift; $s->(@_) }  'u', u($sub, 'foo', 'bar'); 'v', v($sub, 'foo', 'bar');

output:

u: code(0x324718), foo, bar v: foo, bar

i expect u , v behave identically don't. assumed perl evaluated things left right in these situations. code shift->another_method(@_) , shift->another_method(shift, 'stuff', @_) pretty common.

why break if first argument happens code reference? on undefined / undocumented territory here?

the operand evaluation order of ->() undocumented. happens evaluate arguments before lhs (lines 3-4 , 5 respectively below).

>perl -mo=concise,u,-exec a.pl main::u: 1  <;> nextstate(main 51 a.pl:11) v:%,*,&,x*,x&,x$,$,469762048 2  <0> pushmark s 3  <#> gv[*_] s 4  <1> rv2av[t2] lkm/3 5  <0> shift s* 6  <1> entersub[t3] ks/targ,2 7  <1> leavesub[1 ref] k/refc,1 a.pl syntax ok

both using , modifying variable in same expression can dangerous. it's best avoid unless can explain following:

>perl -e"$i=5; $i,++$i,$i" 666

you use

$_[0]->(@_[1..$#_])

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