为什么我无法通过typeglob访问词法变量?
问题描述:
我不明白为什么下面的例子失败(测试2)。为什么我不能从* bb glob访问变量bb?为什么我无法通过typeglob访问词法变量?
use Test::More tests => 4;
$aa = 1;
my $bb = 2; # HERE!
local $cc = 3;
our $dd = 4;
is(${*aa}, 1, "$ \*aa should be 1");
is(${*bb}, 2, "$ \*bb (my) should be 2");
is(${*cc}, 3, "$ \*cc (local) should be 3");
is(${*dd}, 4, "$ \*dd (our) should be 4");
输出是
1..4
ok 1 - *aa should be 1
# Failed test '*bb (my) should be 2'
# at untitled line 10.
# got: undef
# expected: '2'
# Looks like you failed 1 test of 4.
not ok 2 - *bb (my) should be 2
ok 3 - *cc (local) should be 3
ok 4 - *dd (our) should be 4
在perl的5.16.0
由于
怎么样地方? – 2013-02-16 14:12:55
'local'只是给一个全局变量赋值一个临时值。 – choroba 2013-02-16 14:13:16
'local'不会创建变量。 'local'保存一个包变量,散列元素或数组元素的值,并将其恢复到范围退出。 – ikegami 2013-02-16 18:42:46