导入module::类并通过其短名称引用它

Question

当我在Perl 6中导出一个类时，我想只使用它的名称来引用它，而不必指定它的模块:

> use MyLibrary::User;
> User
===SORRY!=== Error while compiling <unknown file>
Undeclared name:
    User used at line 1
> MyLibrary::User;

怎么才能把它的全称叫做MyLibrary::User;而又叫User呢?

Answer 1

你可以很容易地这样做:

constant User = MyLibrary::User;