从 Inline::Perl5 返回的列表给出了项目的计数，而不是列表 [英] list return from Inline::Perl5 gives a count of items, not the list

问题描述

一些简单的 Inline::Perl5 代码返回一个列表，但它似乎返回项目的计数而不是实际列表.

Some simple Inline::Perl5 code returns a list, but it seems to return the count of the items rather than the actual list.

改变涉及的项目数量会改变计数.

Changing the number of items involved changes the count.

use Inline::Perl5;                                             
my $p5 = Inline::Perl5.new;                                    
                                                               
my $perl5_code = q:to/END/;                                    
   sub p5_data {                                               
      my @monsters = qw( godzilla blob tingler kong dorisday );
      return @monsters;                                        
   }                                                           
                                                               
   p5_data();                                                  
END                                                            
                                                               
my @stuff = $p5.run: $perl5_code;                              
say @stuff; # [5]                                              

我以为我会得到存储在数组中的字符串列表，但它的行为就像是将其切换到标量上下文.

I thought I'd get the list of strings stored in the array, but instead it acts like something is switching it to scalar context.

更新:

ikeami 指出返回对的引用效果更好数组:

ikeami points out that it works better to return the reference to the array:

return \@monsters;

不过，你最终会在数组的第一个元素中得到一个数组执行此操作时@stuff 数组:

Though, then you end up with an array in the first element of the @stuff array when you do this:

my @stuff = $p5.run: $perl5_code;                              

另一种方法(来自阅读 Inline::Perl5 文档)是在执行 $p5.run 之后定义 perl5 子，从 perl6 调用它:

An alternate approach (from reading the Inline::Perl5 docs), is after doing a $p5.run to define the perl5 sub, to call it from perl6:

my @stuff = $p5.call('p5_data');  

然后列表返回(return @monsters;)被加载到数组如我所料:

Then the list return (return @monsters;) gets loaded into the array as I expected:

[godzilla blob tingler kong dorisday]

这是最近安装的 0.40 版的 Inline::Perl5，在《乐堂之星 2019.03.1 版构建于 MoarVM 2019.03 版实现Perl 6.d".

This is a recently installed Inline::Perl5 of version 0.40, on "Rakudo Star version 2019.03.1 built on MoarVM version 2019.03 implementing Perl 6.d".

Update2: 所以，似乎运行"暗示标量上下文和调用"是一个列表上下文.

Update2: So, it seems that "run" implies a scalar context and "call" is a list context.

use Inline::Perl5;
my $p5 = Inline::Perl5.new;

my $perl5_defsub = q:to/END/;
   sub whadaya_want {
       wantarray() ? 'LIST' : 'SCALAR';
   }
END

$p5.run: $perl5_defsub;

my $run_context  = $p5.run(  'whadaya_want' );
my $call_context = $p5.call( 'whadaya_want' );  

say "run: ", $run_context;
say "call: ", $call_context;
# run: SCALAR
# call: LIST

推荐答案

Perl5::Inline 将返回值放入标量上下文中.

Perl5::Inline puts return values into scalar context.

作为背景，在 Perl 5 中，上下文向内流入例程，因此例程总是知道它在哪个上下文中.

As background, in Perl 5, context flows inwards into routines, so a routine always knows which context it's in.

在 Perl 6 中，上下文向外流动，因此例程返回一个对象，该对象在不同的​​上下文中表现不同.

In Perl 6, context flows outwards, so a routine returns an object that can behave differently in different context.

Perl 5 和 Perl 6 之间的这种阻抗不匹配意味着 Inline::Perl5 必须决定在一个上下文中调用 Perl 5 例程，这就是标量.

This impedance mismatch between Perl 5 and Perl 6 means that Inline::Perl5 has to decide to call Perl 5 routines in one context, and that's scalar.

正如池上所指出的，正确的解决方案是返回一个正确的标量，也就是引用(在 Perl 5 中).Inline::Perl5 中的限制可能意味着您需要在 Perl 6 端显式取消引用.

As ikegami pointed out, the proper solution is to return a proper scalar, aka reference (in Perl 5 speak). Limitations in Inline::Perl5 might mean you need to explicitly dereference on the Perl 6 side.

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