宏观定义中的“位置太少”

我正在尝试创建一些以这种方式使用（实验）宏的示例：

use experimental :macros;
macro cards_vars() { 
  (<hearts clubs diamonds spades> X~ 1..10).map: { "my \$$^x = False;" }  
};
cards_vars(); 
say $hearts1;

这将创建并运行宏，然后检查定义的变量之一是否存在。但我收到这个错误：

Too few positionals passed; expected 3 arguments but got 2

我什至不知道这个错误来自哪里。我认为它在 cards_vars() 中，但我不知道是否是这样。声明不带括号的宏会产生相同的错误。

Update:需要明确的是，宏不仅是实验性的，而且非常初级且有缺陷。有关此内容的更多信息，请参阅莫里茨对另一个宏问题的回答 https://stackoverflow.com/a/53902789/1077672.

让我们从代码开始：

use experimental :macros;
macro foo { 42 }
foo

这产生相同的编译时 error:

Too few positionals passed; expected 3 arguments but got 2

You get this error if you return anything other than an AST https://docs.perl6.org/type/AST object from a macro.¹ Which is reasonable given that this is the entire point of the macro construct. The error message is less than awesome -- but then again macros are an experimental feature.

所以你需要返回一个 AST。这是一种方法：

use experimental :macros;
macro foo { AST.new }
foo

这产生了run-time error:

Useless use of constant value foo in sink context (line 3)

这表明宏已经完成了它的工作。编译器已完成编译并进入运行时。

明确返回AST对象是错误的处理方式。第一个原因是在AST文档页面：

目前还没有为 AST 定义 API。希望这将成为宏的工作 https://github.com/masak/007.

The second reason is that there's a higher level construct for constructing AST objects, namely the quasi { ... } construct³ mentioned on the doc page and by Scimon:

use experimental :macros;
macro foo { quasi { 42 } }
say foo; # 42

The quasiblock 告诉编译器将所包含的语句编译为 AST 形式。

请注意，结果不是与块对应的 AST。上面它是一个 AST 对象，对应于42.

所以，最后，回到你想要做的事情。

简化：

use experimental :macros;
macro foo { quasi { my $bar } }
foo;
say $bar

yields:

Error while compiling ... Variable '$bar' is not declared

Whereas:

use experimental :macros;
macro foo { quasi { class bar {} } }
foo;
say bar; # (bar)

因此，最重要的是，词法声明目前无法坚持下去。

实验还没有进行到这一步。

Carl Mäsak，Rakudo 中实验宏的创建者，正在继续进行该实验007 https://github.com/masak/007。如果您在 GH 存储库中留下问题，Carl 将会答复。

¹ When the compiler encounters a call to a routine that it knows is a macro² it replaces the macro call with its result. In more detail the compiler:

• 将宏的所有参数编译为 AST 形式；

• 调用宏（在编译时），以 AST 形式传递参数；

• 运行预计返回 AST 对象的宏例程（呃，"ASTish" https://stackoverflow.com/a/53904138/1077672);

• 将生成的 AST(ish) 对象拼接到正在构建的整体 AST 中。

如果您不返回 AST(ish) 对象，那么最后一步就会出错，并出现标题中的错误消息。

² Currently, if you put the macro definition after the call then the compiler doesn't realize it's a macro and treats it as an ordinary routine:

use experimental :macros;
say foo; # 42
macro foo { 42 }
say bar; # AST.new
macro bar { quasi { 42 } }
say bar; # 42

如果这会导致编译时错误，那可能会更好。

³ Carl, the macros champion, likes the word quasi. It draws from lisp heritage which draws from mathematics heritage. Imo it is a "too cute" choice that's a less than awesome fit for P6. My rationale is that Perl philosophy suggests choice of keywords with familiar meaning to newbies. The word "quasi" has a well known normal English connotation of meaning "sort of, but not really". That's not at all helpful. (I have suggested toAST to Carl but he prefers quasi.)