Question

我想知道Modelica是否符合从嵌套的可替换模型扩展模型，即在特定示例中：

package ReplaceableBaseClass
  model ExampleUseReplaceable
    ModelWithReplaceableExtend replaceableExtend1(
      redeclare model LocalModelBase = Extend1,
      input1 = time)
      "Instance with one input and one output";
    ModelWithReplaceableExtend replaceableExtend2(
      redeclare model LocalModelBase = Extend2,
      input1 = Modelica.Math.sin(4*time + 0.3))
      "Instance with one input and two output";
  end ExampleUseReplaceable;

  model ModelWithReplaceableExtend
    "Model which extends from its nested replaceable class"
    extends LocalModelBase;
    replaceable model LocalModelBase = Extend1
      constrainedby PartialToBeExtended;
  end ModelWithReplaceableExtend;

  partial model PartialToBeExtended
    input Real input1;
    output Real output1;
  end PartialToBeExtended;

  model Extend1
    extends PartialToBeExtended;
  equation 
    output1 = 3*input1;
  end Extend1;

  model Extend2
    extends PartialToBeExtended;
    output Real output2;
  equation 
    output1 = input1 + 0.2;
    output2 = input1 * input1;
  end Extend2;
end ReplaceableBaseClass;

这实际上适用于Dymola，但我对它有一种奇怪的感觉，并且不确定使用它是否真的好主意。

Answer 1

这不是合法的Modelica（自Modelica 3.0以来），Dymola将在2017年Dymola之后产生诊断（但只是一个警告）。

Modelica的限制是基类必须是可传递的不可替换的（第7.1.4和6.2.1节）。

从Modelica中的嵌套可替换类扩展类

1 个答案: