perl read（）函数和缓冲区背后的魔力是什么？ [英] What is the magic behind perl read() function and buffer which is not a ref?

问题描述

我无法理解Perl read（$ buf）函数如何能够修改$ buf变量的内容。 $ buf不是引用，因此参数由copy（来自我的c / c ++知识）给出。那么为什么在调用者中修改$ buf变量呢？

I do not get to understand how the Perl read($buf) function is able to modify the content of the $buf variable. $buf is not a reference, so the parameter is given by copy (from my c/c++ knowledge). So how come the $buf variable is modified in the caller ?

这是一个平局变量还是什么？关于setbuf的C文档对我来说也是非常难以捉摸的

Is it a tie variable or something ? The C documentation about setbuf is also quite elusive and unclear to me

# Example 1
$buf=''; # It is a scalar, not a ref
$bytes = $fh->read($buf);
print $buf; # $buf was modified, what is the magic ?

# Example 2
sub read_it {
    my $buf = shift;
    return $fh->read($buf);
}
my $buf;
$bytes = read_it($buf);
print $buf; # As expected, this scope $buf was not modified

推荐答案

不需要魔法 - 如果你愿意，所有perl子程序都是别名调用。 Quoth perlsub ：

No magic is needed -- all perl subroutines are call-by-alias, if you will. Quoth perlsub:

数组@_是一个本地数组，但它的元素是实际标量参数的别名
。特别是，如果更新元素$ _ [0]
，则更新相应的参数（如果不可更新，则发生错误
）。

The array @_ is a local array, but its elements are aliases for the actual scalar parameters. In particular, if an element $_[0] is updated, the corresponding argument is updated (or an error occurs if it is not updatable).

例如：

sub increment {
  $_[0] += 1;
}

my $i = 0;
increment($i);  # now $i == 1

在示例2中，您的 read_it sub 将 @_ 的第一个元素复制到词汇 $ buf ，然后通过调用 read（）就地修改该副本。传入 $ _ [0] 而不是复制，看看会发生什么：

In your "Example 2", your read_it sub copies the first element of @_ to the lexical $buf, which copy is then modified "in place" by the call to read(). Pass in $_[0] instead of copying, and see what happens:

sub read_this {
  $fh->read($_[0]);  # will modify caller's variable
}
sub read_that {
  $fh->read(shift);  # so will this...
}

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