将两个或多个数组传递给Perl子例程

Question

我在子程序中传递和读取参数时遇到问题,该子程序预计会有两个数组.

sub two_array_sum { # two_array_sum ( (1 2 3 4), (2, 4, 0, 1) ) -> (3, 6, 3, 5)
  # I would like to use parameters @a and @b as simply as possible
}

# I would like to call two_array_sum here and pass two arrays, @c and @d

我已经看过并尝试过网上的几个例子,但它们都没有为我工作.

Answer 1

有两种方法可以做到这一点:

1. 原型
2. 引用

但在我讨论这些之前 - 如果你在问题中显示的是关于你想做什么的程度 - 让我建议 List::MoreUtils::pairwise

所以,你要写这个:

my @sum = two_array_sum( @a, @b )

你只需写下:

my @sum = pairwise { $a + $b } @a, @b;

通过原型

这就像push.(就像push它要求@对某些东西有印记一样)

sub two_array_sub (\@\@) { 
    my ( $aref, $bref ) = @_;
    ...
}

那样做的时候

two_array_sub( @a, @b );

有用.通常情况下,它只会在您的子组中显示为一个长列表.它们不适合所有人,正如您在下面的讨论中所看到的那样.

引用

这就是每个人向你展示的方式.

some_sub( \@a, \@b );

关于原型

他们很挑剔.如果您有refs,这将不起作用:

two_array_sub( $arr_ref, $brr_ref );

你必须像这样传递它们:

two_array_sub( @$arr_ref, @$brr_ref );

但是,因为使用深度嵌套的数组使得"数组表达式"变得非常难看,我经常避免Perl的烦躁,因为你可以通过将它放在"字符类"构造中来重载Perl将要采用的引用类型.\[$@]表示引用可以是标量或数组.

sub new_two_array_sub (\[$@]\[$@]) { 
    my $ref = shift;
    my $arr = ref( $ref ) eq 'ARRAY' ? $ref : $$ref; # ref -> 'REF';
    $ref    = shift;
    my $brr = ref( $ref ) eq 'ARRAY' ? $ref : $$ref;
    ...
}

所有这些工作:

new_two_array_sub( @a, $self->{a_level}{an_array} );
new_two_array_sub( $arr, @b );
new_two_array_sub( @a, @b );
new_two_array_sub( $arr, $self->{a_level}{an_array} );

然而,Perl对此仍然很挑剔......出于某种原因:

new_two_array_sub( \@a, $b );
OR 
new_two_array_sub( $a, [ 1..3 ] );

或者任何其他"构造函数"仍然可以被视为对数组的引用.幸运的是,您可以使用旧的Perl 4关闭Perl &

&new_two_array_sub( \@a, [ 1..3 ] );

然后子中的多路复用代码负责处理两个数组引用.

Answer 2

将对数组的引用传递给函数:

two_array_sum( \@a, \@b )

并且不使用a或b作为变量名,因为$a和$b特殊(排序).

Answer 3

我会引用,man perlref但你应该读一遍:

   Making References

   References can be created in several ways.

   1.  By using the backslash operator on a variable, subroutine, or
       value.  (This works much like the & (address-of) operator in C.)
       This typically creates another reference to a variable, because
       there's already a reference to the variable in the symbol table.
       But the symbol table reference might go away, and you'll still have
       the reference that the backslash returned.  Here are some examples:

           $scalarref = \$foo;
           $arrayref  = \@ARGV;
           $hashref   = \%ENV;
           $coderef   = \&handler;
           $globref   = \*foo;

...

   Using References

   That's it for creating references.  By now you're probably dying to
   know how to use references to get back to your long-lost data.  There
   are several basic methods.

   1.  Anywhere you'd put an identifier (or chain of identifiers) as part
       of a variable or subroutine name, you can replace the identifier
       with a simple scalar variable containing a reference of the correct
       type:

           $bar = $$scalarref;
           push(@$arrayref, $filename);
           $$arrayref[0] = "January";
           $$hashref{"KEY"} = "VALUE";
           &$coderef(1,2,3);
           print $globref "output\n";