perl read() 函数和不是引用的缓冲区背后的魔力是什么？

发布于 2024-09-04 15:42:07 字数 496 浏览 9 评论 0原文

我不明白 Perl read($buf) 函数如何修改 $buf 变量的内容。 $buf 不是引用，因此参数是通过 copy 给出的（来自我的 c/c++ 知识）。那么为什么 $buf 变量在调用者中被修改呢？

它是一个领带变量还是什么？关于 setbuf 的 C 文档对我来说也相当难以捉摸且不清楚

# 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

原文

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 ?

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

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梦巷 2024-09-11 15:42:07

不需要任何魔法——如果你愿意的话，所有 Perl 子例程都是通过别名调用的。 perlsub：

数组@_是本地数组，但它的元素是别名
为实际标量参数。特别是，如果一个元素 $_[0]
被更新，对应的参数被更新（或者发生错误
如果它不可更新）。

例如：

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

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

在“示例 2”中，您的 read_it 子将 @_ 的第一个元素复制到词法 $buf< /code>，然后通过调用 read() 来“就地”修改该副本。传入 $_[0] 而不是复制，看看会发生什么：

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

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

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).

For example:

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

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

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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好菇凉咱不稀罕他 2024-09-11 15:42:07

read() 是一个内置函数，因此可以发挥神奇作用。不过，您可以通过声明函数原型，使用自己的函数完成类似的操作：

sub myread(\$) { ... }

参数声明 \$ 表示参数作为引用隐式传递。

内置 read 的唯一魔力是，即使间接调用或作为文件句柄方法调用，它也能工作，而这对常规函数不起作用。

read() is a built-in function, and so can do magic. You can accomplish something similar with your own functions, though, by declaring a function prototype:

sub myread(\$) { ... }

The argument declaration \$ means that the argument is implicitly passed as a reference.

The only magic in the built-in read is that it works even when called indirectly or as a filehandle method, which doesn't work for regular functions.

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