module - Perl6 : use module inside other module

Question

我有 4 个文件。

C:\perlCode2\start.pl6

C:\perlCode2\file0.pm6

C:\perlCode2\folder1\file1.pm6

C:\perlCode2\folder2\file2.pm6

start.pl6 用于运行我的程序。这 3 个模块文件包含或生成最终由 start.pl6 使用的数据。我使用 atom.io 来运行代码。

这是代码:

开始.pl6:

use v6;
use lib ".";
use file0;
use lib "folder1";
use file1;
use lib "folder2";
use file2;

say 'start';
my $file0 = file0.new();
say $file0.mystr;
my $file1 = file1.new();
say $file1.mystr;
my $file2 = file2.new();
say $file2.mystr;
say 'end';

文件 0.pm6:

class file0 is export {
  has Str $.mystr = "file 0";

  submethod BUILD() {
    say "hello file 0";
  }
}

文件 1.pm6:

class file1 is export {
  has Str $.mystr = "file 1";
}

文件 2.pm6:

class file2 is export {
  has Str $.mystr = "file 2";
}

输出:

start
hello file 0
file 0
file 1
file 2
end
[Finished in 0.51s]

我不想在 start.pl6 中创建所有 3 个模块文件的实例，而是在 file1 中创建 file2 的实例，在 file0 中创建 file1 的实例。这样我只需要在 start.pl6 中创建一个 file0 的实例就可以看到相同的输出；

以下是我想到的更改:

文件 1.pm6:

use lib "../folder2";
use "file2.pl6"; 

class file1 is export {
  has Str $.mystr = "file 1";

  submethod BUILD() {
    my $file2 = file2.new();
    $!mystr = $!mystr ~ "\n" ~ $file2.mystr; 
        # I want to instantiate file2 inside the constructor, 
        # so I can be sure the line
        # $!mystr = $!mystr ~ "\n" ~ $file2.mystr; 
        # takes effect before i call any of file0's methods;
  }
}

文件 0.pm6:

use lib "folder1";
use "file1.pl6"; 

class file0 is export {
  has Str $.mystr = "file 0";

  submethod BUILD() {
    say "hello file 0";
    my $file1 = file1.new();
    $!mystr = $!mystr ~ "\n" ~ $file1.mystr; 
  }
}

在 file0 中，行
使用库“文件夹1”；
使用“file1.pl6”；
产生此错误:

===SORRY!=== Error while compiling C:\perlCode2\file0.pm6 (file0)
'use lib' may not be pre-compiled
at C:\perlCode2\file0.pm6 (file0):2
------> use lib "folder1/file1.pl6"<HERE>;
[Finished in 0.584s]

我file1，行
使用库“../folder2”；
使用“文件2”；
不起作用，但也没有给出错误。我只是得到输出:
【0.31s完成】

最后，文件 start.pl6 应该如下所示以产生输出:

开始.pl6:

use v6;
use lib ".";
use file0;

say 'start';
my $file0 = file0.new();
say $file0.mystr;
say 'end';

输出:

start
hello file 0
file 0
file 1
file 2
end

Answer 1

你试图做的事情对我来说毫无意义。您似乎将这些模块任意放在文件夹中。

如果这些模块的名称确实有意义，那么我可能会如何构建它。

C:\perlCode2\start.pl6
C:\perlCode2\lib\file0.pm6
C:\perlCode2\lib\folder1\file1.pm6
C:\perlCode2\lib\folder2\file2.pm6

开始.pl6:

use v6;

END say "[Finished in {(now - $*INIT-INSTANT).fmt("%0.2fs")}";

use lib 'lib';
use file0;

say 'start';
my $file0 = file0.new;
say $file0.mystr;
say 'end';

lib\file0.pm6:

use folder1::file1;

class file0 is export {
  has Str $.mystr = "file 0";

  submethod TWEAK() {
    say "hello file 0";
    $!mystr ~= "\n" ~ folder1::file1.new.mystr; 
  }
}

lib\folder1\file1.pm6:

use folder2::file2;

class folder1::file1 is export {
  has Str $.mystr = "file 1";

  submethod TWEAK() {
    $!mystr ~= "\n" ~ folder2::file2.new.mystr;
  }
}

lib\folder2\file2.pm6

class folder2::file2 is export {
  has Str $.mystr = "file 2";
}