Question

以下是我的代码

sub print_list {
  $max = $_[0];
  for ($i=0; $i<$max; $i++)
  {
    print "$i.  $list[$i][0]\t $list[$i][1]\n";
  }
}

# Declaring a 2-D Array, which is just an array of 1-D arrays

@list = ( ["vi   ", "Null"], ["emacs", "Null"], ["joe  ", "Null" ]);

$max = $#list + 1;

print "Initial Values\n";
print_list($max);

print "\n\n";

输出： -

 Initial Values 
 0. vi                 Null 
 1. emacs           Null 
 2. joe               Null

在上面的代码中，$ max传递给函数的方法如何获得函数中的所有可用数据。

Answer 1

将$ max传递给函数也不会将@list数组传递给函数。通过什么来证明这一点：

sub print_list {
  for ($i=0; $i<$max; $i++)
  {
    print "$i.  $list[$i][0]\t $list[$i][1]\n";
  }
}

# Declaring a 2-D Array, which is just an array of 1-D arrays

@list = ( ["vi   ", "Null"], ["emacs", "Null"], ["joe  ", "Null" ]);

$max = $#list + 1;

print "Initial Values\n";
print_list();

print "\n\n";

@list数组变量是全局的。

perldoc perlsub

Answer 2

@list数据根本没有传递给子程序。您通过在子例程之外可见的@list版本访问它。访问未传递到子例程中的变量是一个非常糟糕的主意，因为您的子例程与调用它的代码变得过于紧密耦合。例如，在您的代码中，print_list只能处理名为@list的变量。如果将数组传递给子例程，则其名称无关紧要。

# I've renamed this as you're dealing with arrays, not lists
sub print_array {
  my $length = shift;
  my @array  = @_;

  for ($i = 0; $i < $max; $i++) {
    print "$i.  $array[$i][0]\t $array[$i][1]\n";
  }
}

my @editors = ( ["vi   ", "Null"], ["emacs", "Null"], ["joe  ", "Null" ]);
my $number_of_editors = @editors; # More readable than $#editors + 1

print_array($number_of_editors, @editors);

但是你根本不需要将数组长度传递给子程序。

# I've renamed this as you're dealing with arrays, not lists
sub print_array {
  my @array  = @_;

  for ($i = 0; $i < @array; $i++) {
    print "$i.  $array[$i][0]\t $array[$i][1]\n";
  }
}

my @editors = ( ["vi   ", "Null"], ["emacs", "Null"], ["joe  ", "Null" ]);

print_array(@editors);

作为最后的改进，{C foreach循环几乎总是更容易遵循for循环

# I've renamed this as you're dealing with arrays, not lists
sub print_array {
  my @array  = @_;

  foreach (0 .. $#array) {
    print "$_.  $array[$_][0]\t $array[$_][1]\n";
  }
}

Perl Reference，传递计数变量（$＃）如何获得完整数据的引用？

2 个答案: