如何在 Chapel 中保留可接受类型的列表以进行比较

How to keep a list of acceptable types for comparison in Chapel

假设我有 class StudentBadStudent:StudentGoodStudent:StudentExcellentStudent: Student。我想要一个 class 方法只对 GoodExceptional 学生进行操作。类似于:

class AdvancedBasketWeaving {

  // this is the question:
  var acceptableStudentTypes: [1..2] = [GoodStudent, ExcellentStudent];

  proc accept(student: Student) {
    for at in this.acceptableStudentTypes {
      if student.type == at then return "YES!";
    }
    return "Computer says 'No'";
  }
}

如何获得此功能?

您可以尝试将类型转换为 string

例如,

var a = [int: string, string: string];
writeln(a);
var b = 1;
var c = "sdasas";
if b.type: string == a[1] then writeln("This matches!");
if c.type: string != a[1] then writeln("This doesn't match!");

我认为您会希望将两种工具用于此模式:

1) 首先是 Chapel 的 cast 运算符 (:)。对于 classes,转换类似于 C++ 的动态转换。简而言之,给定一个 class 对象,例如 GoodStudent 的以下实例:

var brian = new GoodStudent();

将对象转换为 class 类型将 return nil 如果对象不是 class 的子class 并且class参考如果是。因此:

...(brian: Student != nil)...           // will evaluate to true
...(brian: BadStudent != nil)...        // will evaluate to false
...(brian: GoodStudent != nil)...       // will evaluate to true
...(brian: ExcellentStudent != nil)...  // will evaluate to false

因此,要测试 GoodStudentExcellentStudent,您可以这样写:

if (student:GoodStudent != nil || student:ExcellentStudent != nil) then
  return "YES!";

或者,如果每个 ExcellentStudent 也是一个 GoodStudent,您可以考虑将其设为 class 层次结构中 GoodStudent 的子 class 而不是比它的兄弟姐妹。在这种情况下,您可以简单地将条件写为:

if student:GoodStudent != nil then return "YES!";

因为 GoodStudentExcellentStudent 都会 return 为真。

作为重要说明,可能很容易将此条件写为:

if student.type == GoodStudent

但这不会在您的过程上下文中给出正确的行为,因为它声明如下:

proc accept(student: Student) { ... }

具体来说,.type 查询将 return class 对象的静态(编译时)类型,并且在此例程的上下文中,studentStudent 由于它的正式类型。所以比较它的静态类型永远不会匹配 GoodStudent,即使对象的动态类型是 GoodStudent。使用动态转换通过将静态测试更改为动态测试来解决此问题。另一种方法是使 accept() 过程完全通用,如下所示:

proc accept(student) { ... }

但是随后您通过允许传入其他非Student类型来打开闸门。

2) 您需要的第二件事(也是您问题的重点)是元组类型,这可能是创建类型集合的最佳/最轻量级方法。 Chapel 仅支持值数组,不支持类型,因此代码中的以下行不合法:

var acceptableStudentTypes: [1..2] = [GoodStudent, ExcellentStudent];

相反,创建一个元组类型来存储您要比较的所有类型:

type acceptableStudentTypes = (GoodStudent, ExcellentStudent);

这引出了我建议的解决方案 (try it online):

class Student {
}

class BadStudent: Student {
}

class GoodStudent: Student {
}

class ExcellentStudent: Student {
}

// a tuple of acceptable types                                              
type acceptableStudentTypes = (GoodStudent, ExcellentStudent);

class AdvancedBasketWeaving {
  proc accept(student: Student) {
    // iterate over the tuple's size                                        
    for param i in 1..acceptableStudentTypes.size do
      // see if dynamically casting the student to the type "works"         
      if student: acceptableStudentTypes(i) != nil then
        return "YES!";
    return "Computer says 'No'";
  }
}

var course = new AdvancedBasketWeaving();
writeln(course.accept(new Student()));            // Computer says 'No'     
writeln(course.accept(new BadStudent()));         // Computer says 'No'     
writeln(course.accept(new GoodStudent()));        // YES!                   
writeln(course.accept(new ExcellentStudent()));   // YES!                   

请注意,我已将 acceptableStudentTypes 声明从 class 范围(这是合乎逻辑的,您拥有它的位置)移至模块范围。这是因为我 filed an issue against.

Chapel 中的一个明显错误

或者,如果您可以使 ExcellentStudent 成为 GoodStudent 的子 class,我认为以下内容更好 (try it online):

class Student {
}

class BadStudent: Student {
}

class GoodStudent: Student {
}

class ExcellentStudent: GoodStudent {
}

class AdvancedBasketWeaving {
  proc accept(student: Student) {
    if student: GoodStudent != nil then
      return "YES!";
    return "Computer says 'No'";
  }
}

var course = new AdvancedBasketWeaving();
writeln(course.accept(new Student()));            // Computer says 'No'     
writeln(course.accept(new BadStudent()));         // Computer says 'No'     
writeln(course.accept(new GoodStudent()));        // YES!                   
writeln(course.accept(new ExcellentStudent()));   // YES!