perl read()函数和缓冲区背后的魔力是什么？

我不明白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？

它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理解的

# 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

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

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

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