数组和哈希作为状态变量有限制.在Perl 5.10中,我们无法在列表上下文中初始化它们：
所以
state @array = qw(a b c); #Error!
为什么会这样？为什么不允许？
我们可以使用状态数组,并通过这种方式初始化它们
state @numbers; push @numbers,5; push @numbers,6;
但是为什么不直接用state @numbers = qw(5 6)呢？
为什么Perl不允许
解决方法
根据
perldiag,计划在未来版本中支持列表上下文初始化：
- Initialization of state variables in list context currently forbidden
(F) Currently the implementation of “state” only permits the initialization of scalar variables in scalar context. Re-writestate ($a) = 42asstate $a = 42to change from list to scalar context. Constructions such asstate (@a) = foo()will be supported in a future perl release.
根据this message about the change that made this an error：
For now,forbid all list assignment initialisation of state variables,
as the precise semantics in Perl 6 are not clear. Better to make it a
Syntax error,than to have one behavIoUr now,but change it later.
[I believe that this is the consensus. If not,it will be backed out]
你可以随时使用arrayref：
state $arrayRef = [qw(a b c)];
注意你的例子
state @numbers; push @numbers,6;
并不意味着状态@numbers = qw(5 6)将(如果它工作)相同的事情.状态变量只能初始化一次,但是你的代码会推送5&每次执行该代码时,都会将该数组写入数组.