Mixing
Using alternative super class constructor in child class instantiation
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I have a base class with two constructors, and a child class which has one constructor. Is it possible to instantiate a child class using the second base class constructor?
Example code:
abstract class RuleCondition(rule:Rule, field:String, equal:Boolean, inverted:Boolean)
{
// alternate constructor with RuleValue instead of static comparation value
def this(rule:Rule, field:String, ref:RuleValue, equal:Boolean = false, inverted:Boolean = false) = ???
}
class RuleConditionAbove(rule:Rule, field:String, comparationValue:Long, equal:Boolean = false, inverted:Boolean = false)
extends RuleCondition(rule, field, equal, inverted)
{
// ...
}
Now I can do this:
val myAboveCondition = new RuleConditionAbove(rule, "bla", 10, true, false)
but I cannot do this:
val myAboveCondition = new RuleConditionAbove(rule, "bla", RuleValue(...), true, false)
because the alternative constructor of RuleCondition base class is not visible. It will be visible once I add this to the child class:
def this(rule:Rule, field:String, ref:RuleValue, equal:Boolean = false, inverted:Boolean = false) = this(rule, field, ref, equal, inverted)
Would this be the only/usual way of solving this issue, or is there something smarter which involves less copy & past code? (since I have a ton of child classes of the same pattern)
[edit] To clarify, the second constructor would be the same in every child class, thus I would like to have it implemented only once in the base class.
However still having to put another constructor in each child class would defeat this purpose somehow, and thus I would not have two constructors in the base class but rather only in all child classes.
scala inheritance constructor polymorphism superclass
edited Jan 3 at 11:23
Jeffrey Chung
14.6k62242
asked Jan 3 at 8:13
user826955user826955
1,24711740
add a comment |
I have a base class with two constructors, and a child class which has one constructor. Is it possible to instantiate a child class using the second base class constructor?
Example code:
abstract class RuleCondition(rule:Rule, field:String, equal:Boolean, inverted:Boolean)
{
// alternate constructor with RuleValue instead of static comparation value
def this(rule:Rule, field:String, ref:RuleValue, equal:Boolean = false, inverted:Boolean = false) = ???
}
class RuleConditionAbove(rule:Rule, field:String, comparationValue:Long, equal:Boolean = false, inverted:Boolean = false)
extends RuleCondition(rule, field, equal, inverted)
{
// ...
}
Now I can do this:
val myAboveCondition = new RuleConditionAbove(rule, "bla", 10, true, false)
but I cannot do this:
val myAboveCondition = new RuleConditionAbove(rule, "bla", RuleValue(...), true, false)
because the alternative constructor of RuleCondition base class is not visible. It will be visible once I add this to the child class:
def this(rule:Rule, field:String, ref:RuleValue, equal:Boolean = false, inverted:Boolean = false) = this(rule, field, ref, equal, inverted)
Would this be the only/usual way of solving this issue, or is there something smarter which involves less copy & past code? (since I have a ton of child classes of the same pattern)
[edit] To clarify, the second constructor would be the same in every child class, thus I would like to have it implemented only once in the base class.
However still having to put another constructor in each child class would defeat this purpose somehow, and thus I would not have two constructors in the base class but rather only in all child classes.
scala inheritance constructor polymorphism superclass
edited Jan 3 at 11:23
Jeffrey Chung
14.6k62242
asked Jan 3 at 8:13
user826955user826955
1,24711740
add a comment |
I have a base class with two constructors, and a child class which has one constructor. Is it possible to instantiate a child class using the second base class constructor?
Example code:
abstract class RuleCondition(rule:Rule, field:String, equal:Boolean, inverted:Boolean)
{
// alternate constructor with RuleValue instead of static comparation value
def this(rule:Rule, field:String, ref:RuleValue, equal:Boolean = false, inverted:Boolean = false) = ???
}
class RuleConditionAbove(rule:Rule, field:String, comparationValue:Long, equal:Boolean = false, inverted:Boolean = false)
extends RuleCondition(rule, field, equal, inverted)
{
// ...
}
Now I can do this:
val myAboveCondition = new RuleConditionAbove(rule, "bla", 10, true, false)
but I cannot do this:
val myAboveCondition = new RuleConditionAbove(rule, "bla", RuleValue(...), true, false)
because the alternative constructor of RuleCondition base class is not visible. It will be visible once I add this to the child class:
def this(rule:Rule, field:String, ref:RuleValue, equal:Boolean = false, inverted:Boolean = false) = this(rule, field, ref, equal, inverted)
Would this be the only/usual way of solving this issue, or is there something smarter which involves less copy & past code? (since I have a ton of child classes of the same pattern)
[edit] To clarify, the second constructor would be the same in every child class, thus I would like to have it implemented only once in the base class.
However still having to put another constructor in each child class would defeat this purpose somehow, and thus I would not have two constructors in the base class but rather only in all child classes.
scala inheritance constructor polymorphism superclass
edited Jan 3 at 11:23
Jeffrey Chung
14.6k62242
asked Jan 3 at 8:13
user826955user826955
1,24711740
I have a base class with two constructors, and a child class which has one constructor. Is it possible to instantiate a child class using the second base class constructor?
Example code:
abstract class RuleCondition(rule:Rule, field:String, equal:Boolean, inverted:Boolean)
{
// alternate constructor with RuleValue instead of static comparation value
def this(rule:Rule, field:String, ref:RuleValue, equal:Boolean = false, inverted:Boolean = false) = ???
}
class RuleConditionAbove(rule:Rule, field:String, comparationValue:Long, equal:Boolean = false, inverted:Boolean = false)
extends RuleCondition(rule, field, equal, inverted)
{
// ...
}
Now I can do this:
val myAboveCondition = new RuleConditionAbove(rule, "bla", 10, true, false)
but I cannot do this:
val myAboveCondition = new RuleConditionAbove(rule, "bla", RuleValue(...), true, false)
because the alternative constructor of RuleCondition base class is not visible. It will be visible once I add this to the child class:
def this(rule:Rule, field:String, ref:RuleValue, equal:Boolean = false, inverted:Boolean = false) = this(rule, field, ref, equal, inverted)
Would this be the only/usual way of solving this issue, or is there something smarter which involves less copy & past code? (since I have a ton of child classes of the same pattern)
[edit] To clarify, the second constructor would be the same in every child class, thus I would like to have it implemented only once in the base class.
However still having to put another constructor in each child class would defeat this purpose somehow, and thus I would not have two constructors in the base class but rather only in all child classes.
scala inheritance constructor polymorphism superclass
scala inheritance constructor polymorphism superclass
edited Jan 3 at 11:23
Jeffrey Chung
14.6k62242
asked Jan 3 at 8:13
user826955user826955
1,24711740
edited Jan 3 at 11:23
Jeffrey Chung
14.6k62242
asked Jan 3 at 8:13
user826955user826955
1,24711740
edited Jan 3 at 11:23
Jeffrey Chung
14.6k62242
edited Jan 3 at 11:23
Jeffrey Chung
14.6k62242
edited Jan 3 at 11:23
Jeffrey Chung
14.6k62242
14.6k62242
asked Jan 3 at 8:13
user826955user826955
1,24711740
asked Jan 3 at 8:13
user826955user826955
1,24711740
asked Jan 3 at 8:13
user826955user826955
1,24711740
1,24711740
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
Is it possible to instanciate a child class using the [second] base class constructor?
No.
You can never use a superclass constructor to create an instance of a subclass. You have to call a constructor for the class that you are creating. The subclass constructor must call a constructor for the superclass, but you can't call it directly.
So the reason that you can do this
val myAboveCondition = new RuleConditionAbove(rule, "bla", 10, true, false)
is that RuleConditionAbove has a constructor with those arguments. It has nothing to do with the fact that RuleCondition has a constructor with the same arguments.
And the reason that you can't do this
val myAboveCondition = new RuleConditionAbove(rule, "bla", RuleValue(...), true, false)
is that RuleConditionAbove does not have a constructor with those arguments.
answered Jan 3 at 8:49
TimTim
6,8071819
add a comment |
You have to add a constructor definition in every child class as you described.
def this(rule:Rule, field:String, ref:RuleValue, equal:Boolean = false, inverted:Boolean = false) = this(rule, field, ref, equal, inverted)
Imagine that a child class defines new fields that are not available in the base class. Creating child class with a base constructor would not define such fields and leave the instance of the class partially initialised.
If your base constructor has valuable logic, it makes sense to keep it in the base class. And just "link" it to the base constructor in the child class.
answered Jan 3 at 8:52
Ivan StanislavciucIvan Stanislavciuc
1,670411
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Is it possible to instanciate a child class using the [second] base class constructor?
No.
You can never use a superclass constructor to create an instance of a subclass. You have to call a constructor for the class that you are creating. The subclass constructor must call a constructor for the superclass, but you can't call it directly.
So the reason that you can do this
val myAboveCondition = new RuleConditionAbove(rule, "bla", 10, true, false)
is that RuleConditionAbove has a constructor with those arguments. It has nothing to do with the fact that RuleCondition has a constructor with the same arguments.
And the reason that you can't do this
val myAboveCondition = new RuleConditionAbove(rule, "bla", RuleValue(...), true, false)
is that RuleConditionAbove does not have a constructor with those arguments.
answered Jan 3 at 8:49
TimTim
6,8071819
add a comment |
Is it possible to instanciate a child class using the [second] base class constructor?
No.
You can never use a superclass constructor to create an instance of a subclass. You have to call a constructor for the class that you are creating. The subclass constructor must call a constructor for the superclass, but you can't call it directly.
So the reason that you can do this
val myAboveCondition = new RuleConditionAbove(rule, "bla", 10, true, false)
is that RuleConditionAbove has a constructor with those arguments. It has nothing to do with the fact that RuleCondition has a constructor with the same arguments.
And the reason that you can't do this
val myAboveCondition = new RuleConditionAbove(rule, "bla", RuleValue(...), true, false)
is that RuleConditionAbove does not have a constructor with those arguments.
answered Jan 3 at 8:49
TimTim
6,8071819
add a comment |
Is it possible to instanciate a child class using the [second] base class constructor?
No.
You can never use a superclass constructor to create an instance of a subclass. You have to call a constructor for the class that you are creating. The subclass constructor must call a constructor for the superclass, but you can't call it directly.
So the reason that you can do this
val myAboveCondition = new RuleConditionAbove(rule, "bla", 10, true, false)
is that RuleConditionAbove has a constructor with those arguments. It has nothing to do with the fact that RuleCondition has a constructor with the same arguments.
And the reason that you can't do this
val myAboveCondition = new RuleConditionAbove(rule, "bla", RuleValue(...), true, false)
is that RuleConditionAbove does not have a constructor with those arguments.
answered Jan 3 at 8:49
TimTim
6,8071819
Is it possible to instanciate a child class using the [second] base class constructor?
No.
You can never use a superclass constructor to create an instance of a subclass. You have to call a constructor for the class that you are creating. The subclass constructor must call a constructor for the superclass, but you can't call it directly.
So the reason that you can do this
val myAboveCondition = new RuleConditionAbove(rule, "bla", 10, true, false)
is that RuleConditionAbove has a constructor with those arguments. It has nothing to do with the fact that RuleCondition has a constructor with the same arguments.
And the reason that you can't do this
val myAboveCondition = new RuleConditionAbove(rule, "bla", RuleValue(...), true, false)
is that RuleConditionAbove does not have a constructor with those arguments.
answered Jan 3 at 8:49
TimTim
6,8071819
answered Jan 3 at 8:49
TimTim
6,8071819
answered Jan 3 at 8:49
TimTim
6,8071819
answered Jan 3 at 8:49
TimTim
6,8071819
6,8071819
add a comment |
add a comment |
You have to add a constructor definition in every child class as you described.
def this(rule:Rule, field:String, ref:RuleValue, equal:Boolean = false, inverted:Boolean = false) = this(rule, field, ref, equal, inverted)
Imagine that a child class defines new fields that are not available in the base class. Creating child class with a base constructor would not define such fields and leave the instance of the class partially initialised.
If your base constructor has valuable logic, it makes sense to keep it in the base class. And just "link" it to the base constructor in the child class.
answered Jan 3 at 8:52
Ivan StanislavciucIvan Stanislavciuc
1,670411
add a comment |
You have to add a constructor definition in every child class as you described.
def this(rule:Rule, field:String, ref:RuleValue, equal:Boolean = false, inverted:Boolean = false) = this(rule, field, ref, equal, inverted)
Imagine that a child class defines new fields that are not available in the base class. Creating child class with a base constructor would not define such fields and leave the instance of the class partially initialised.
If your base constructor has valuable logic, it makes sense to keep it in the base class. And just "link" it to the base constructor in the child class.
answered Jan 3 at 8:52
Ivan StanislavciucIvan Stanislavciuc
1,670411
add a comment |
You have to add a constructor definition in every child class as you described.
def this(rule:Rule, field:String, ref:RuleValue, equal:Boolean = false, inverted:Boolean = false) = this(rule, field, ref, equal, inverted)
Imagine that a child class defines new fields that are not available in the base class. Creating child class with a base constructor would not define such fields and leave the instance of the class partially initialised.
If your base constructor has valuable logic, it makes sense to keep it in the base class. And just "link" it to the base constructor in the child class.
answered Jan 3 at 8:52
Ivan StanislavciucIvan Stanislavciuc
1,670411
You have to add a constructor definition in every child class as you described.
def this(rule:Rule, field:String, ref:RuleValue, equal:Boolean = false, inverted:Boolean = false) = this(rule, field, ref, equal, inverted)
Imagine that a child class defines new fields that are not available in the base class. Creating child class with a base constructor would not define such fields and leave the instance of the class partially initialised.
If your base constructor has valuable logic, it makes sense to keep it in the base class. And just "link" it to the base constructor in the child class.
answered Jan 3 at 8:52
Ivan StanislavciucIvan Stanislavciuc
1,670411
answered Jan 3 at 8:52
Ivan StanislavciucIvan Stanislavciuc
1,670411
answered Jan 3 at 8:52
Ivan StanislavciucIvan Stanislavciuc
1,670411
answered Jan 3 at 8:52
Ivan StanislavciucIvan Stanislavciuc
1,670411
1,670411
add a comment |
add a comment |
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