Question

我在另一篇文章的回答中看到了这段代码：Why would I use Perl anonymous subroutines instead of a named one?，但无法确切知道发生了什么，所以我想自己运行它。

sub outer
{
  my $a = 123;

  sub inner
  {
    print $a, "\n"; #line 15 (for your reference, all other comments are the OP's)
  }

  # At this point, $a is 123, so this call should always print 123, right?
  inner();

  $a = 456;
}

outer(); # prints 123
outer(); # prints 456! Surprise!

在上面的例子中，我收到一条警告：“变量$ a不会在第15行保持共享。显然，这就是输出“意外”的原因，但我仍然不明白这里发生了什么。

sub outer2
{
  my $a = 123;

  my $inner = sub
  {
    print $a, "\n";
  };

  # At this point, $a is 123, and since the anonymous subrotine 
  # whose reference is stored in $inner closes over $a in the 
  # "expected" way...
  $inner->();

  $a = 456;
}

# ...we see the "expected" results
outer2(); # prints 123
outer2(); # prints 123

同样，我也不明白这个例子中发生了什么。有人可以解释一下吗？

提前致谢。

Answer 1

它与子例程的编译时与运行时解析有关。正如diagnostics消息所示，

当调用内部子程序时，它将看到的值外部子程序的变量与第一个之前和期间的变量一样调用外部子程序;在这种情况下，第一次调用之后外子程序完成后，内子程序和外子程序都没有更长时间共享变量的公共值。换句话说，变量将不再被共享。

注释您的代码：

sub outer
{
  # 'my' will reallocate memory for the scalar variable $a
  # every time the 'outer' function is called. That is, the address of
  # '$a' will be different in the second call to 'outer' than the first call.

  my $a = 123;


  # the construction 'sub NAME BLOCK' defines a subroutine once,
  # at compile-time.

  sub inner1
  {

    # since this subroutine is only getting compiled once, the '$a' below
    # refers to the '$a' that is allocated the first time 'outer' is called

    print "inner1: ",$a, "\t", \$a, "\n"; 
  }

  # the construction  sub BLOCK  defines an anonymous subroutine, at run time
  # '$inner2' is redefined in every call to 'outer'

  my $inner2 = sub {

    # this '$a' now refers to '$a' from the current call to outer

    print "inner2: ", $a, "\t", \$a, "\n";
  };

  # At this point, $a is 123, so this call should always print 123, right?
  inner1();
  $inner2->();

  # if this is the first call to 'outer', the definition of 'inner1' still
  # holds a reference to this instance of the variable '$a', and this
  # variable's memory will not be freed when the subroutine ends.

  $a = 456;
}
outer();
outer();

典型输出：

inner1: 123     SCALAR(0x80071f50)
inner2: 123     SCALAR(0x80071f50)
inner1: 456     SCALAR(0x80071f50)
inner2: 123     SCALAR(0x8002bcc8)

Answer 2

你可以打印\＆amp; inner;在第一个例子中（定义后），并打印$ inner;在第二。

您看到的是十六进制代码引用，在第一个示例中相等，在第二个示例中有所不同。所以，在第一个例子中，内部只创建了一次，并且它总是从外部（）的第一次调用中封闭到$一个词法变量。

子例程与匿名子例程的上下文中的共享变量

2 个答案: