您不必傳遞參考$i
。 $_[0]
是別名的$i
,當您調用它作爲subr($i)
時。
use strict;
use warnings;
use Test::More tests => 2;
sub subr{ $_[0]++ } # messing with exactly what was passed first
my $i=2;
is($i, 2, q[$i == 2]);
subr($i);
is($i, 3, q[$i == 3]);
又如是這樣的:
use strict;
use warnings;
use Test::More tests => 6;
use Test::Exception;
sub subr{ $_[0]++ }
my $i=2;
is($i, 2, q[$i == 2]);
subr($i);
is($i, 3, q[$i == 3]);
sub subr2 { $_[0] .= 'x'; }
dies_ok { subr2('lit'); } 'subr2 *dies* trying to modify a literal';
lives_ok {
my $s = 'lit';
subr2($s);
is($s, 'litx', q[$s eq 'litx']);
subr2((my $s2 = 'lit'));
is($s2, 'litx', q[$s2 eq 'litx']);
} 'subr2 lives with heap variables';
輸出:
ok 1 - $i == 2
ok 2 - $i == 3
ok 3 - subr2 *dies* trying to modify a literal
ok 4 - $s eq 'litx'
ok 5 - $s2 eq 'litx'
ok 6 - subr2 lives with heap variables
1..6
這太好了,謝謝 – user685275 2011-05-03 16:26:45