当我在Perl 6中导出类时,我只想使用其名称来引用它,而不必指定其模块:
> use MyLibrary::User;
> User
===SORRY!=== Error while compiling <unknown file>
Undeclared name:
User used at line 1
> MyLibrary::User;
以全名MyLibrary::User;
调用并能够以User
调用的方式是什么?
您可以很容易地做到这一点:
constant User = MyLibrary::User;